Transcript 3.1 Solving Linear Equations Part I • A linear equation in one variable can be written in the form: Ax + B.

3.1 Solving Linear Equations Part I
• A linear equation in one variable can be written in
the form: Ax + B = 0
• Linear equations are solved by getting “x” by
itself on one side of the equation
• Addition (Subtraction) Property of Equality:
if a  b, then
a  c  b  c and
ac  bc
3.1 Solving Linear Equations Part I
• Multiplication Property of Equality:
if a  b, c  0 then
ac  bc
• Since division is the same as multiplying by the reciprocal,
you can also divide each side by a number.
if a  b, c  0 then
a b

c c
• General rule: Whatever you do to one side of the equation,
you have to do the same thing to the other side.
3.1 Solving Linear Equations Part I
• Example: Solve by getting x by itself on one side of
the equation.
Subtract 7 from both sides:
Divide both sides by 3:
3x  7  52
3x  45
x  15
3.2 Solving Linear Equations Part II
- Fractions/Decimals
• As with expressions, you need to combine like
terms and use the distributive property in equations.
Example:
3 x  5 x  7(15  7)
8 x  105 49  56
x7
3.2 Solving Linear Equations Part II
- Fractions/Decimals
• Fractions - Multiply each term on both sides by the Least
Common Denominator (in this case the LCD = 4):
1
 x  5  1 x  3
4
2
Reduce Fractions: 4  x  5  4 x  4  3
4
2
 x  5  2 x  1 2
Subtract x:
5  x 12
Subtract 5:
x  17
Multiply by 4:
3.2 Solving Linear Equations Part II
- Fractions/Decimals
• Decimals - Multiply each term on both sides by the
smallest power of 10 that gets rid of all the decimals
Multiply by 100:
Cancel:
Distribute:
Subtract 5x:
Subtract 50:
Divide by 5:
.1 x  5  .05x  .3
100 .1x  5  100 .05x  100 .3
10x  5  5 x  30
10x  50  5 x  30
5 x  50  30
5 x  80  x  16
3.2 Solving Linear Equations Part II
- Fractions/Decimals
• Eliminating fractions makes the calculation simpler:
1
1
 x  5 
x 1
Multiply by 94:
47
94
94
94
Cancel:
 x  5 
x  9 4 1
47
94
2 x  5   x  9 4
Distribute:
2x 10  x  94
Subtract x:
x  1 0  9 4
Subtract 10:
x  1 04
3.2 Solving Linear Equations Part II
• 1 – Multiply on both sides to get rid of
fractions/decimals
• 2 – Use the distributive property
• 3 – Combine like terms
• 4 – Put variables on one side, numbers on the
other by adding/subtracting on both sides
• 5 – Get “x” by itself on one side by multiplying or
dividing on both sides
• 6 – Check your answers (if you have time)
3.2 Solving Linear Equations Part II
• Example:
Clear fractions:
2
3
x  x  x 3
Combine like terms:
4 x  3 x  x  18
1
2
1
6
Get variables on one side:
7 x  x  18
Solve for x:
6 x  18
x3
3.3 Applications of Linear Equations to
General Problems
• 1 – Decide what you are asked to find
• 2 – Write down any other pertinent information
(use other variables, draw figures or diagrams )
• 3 – Translate the problem into an equation.
• 4 – Solve the equation.
• 5 – Answer the question posed.
• 6 – Check the solution.
3.3 Applications of Linear Equations to
General Problems
• Example: The sum of 3 consecutive integers is
126. What are the integers?
x = first integer, x + 1 = second integer,
x + 2 = third integer
x  ( x  1)  ( x  2)  126
3 x  3  126
3 x  123
x  41
41, 42, 43
3.3 Applications of Linear Equations to
General Problems
• Example: Renting a car for one day costs $20 plus $.25
per mile. How much would it cost to rent the car for
one day if 68 miles are driven?
$20 = fixed cost,
$.25  68 = variable cost
$20  68 $.25 
$20  $17 
$37
3.4 Percent Increase/Decrease
and Investment Problems
• A number increases from 60 to 81. Find the
percent increase in the number.
increase  81 60  21
21
% increase 
 0.35
60
 35%
3.4 Percent Increase/Decrease
and Investment Problems
• A number decreases from 81 to 60. Find the
percent increase in the number.
decrease 81 60  21
21
% decrease
 0.26
81
 26%
Why is this percent different than the last slide?
3.4 Percent Increase/Decrease
and Investment Problems
• A flash drive is on sale for $12 after a 20%
discount. What was the original price of the flash
drive?
x  20% x  12
1.0 x  .2 x  12
.8 x  12
x  12
.8  $15
3.4 Percent Increase/Decrease
and Investment Problems
• Another Way: A flash drive is on sale for $12 after a 20%
discount. What was the original price of the flash drive?
Since $12 was on sale for 20% off, it is 100% - 20% =
80% of the original price set up as a proportion (see 3.6):
12
80

x
100
80x  12(100)  1200
1200
x
 15
80
3.4 Percent Increase/Decrease
and Investment Problems
• Simple Interest Formula: I  PRT
I = interest
P = principal
R = rate of interest per year
T = time in years
3.4 Percent Increase/Decrease
and Investment Problems
• Example: Given an investment of $9500
invested at 12% interest for 1½ years, find
the simple interest.
I  PRT  $9500 0.12  1.5
I  $1710
3.4 Percent Increase/Decrease
and Investment Problems
• Example: If money invested at 10% interest
for 2 years yields $84, find the principal.
I  PRT
$84  P  0.10 2  .2 P
$84
P
 $420
.2
3.5 Geometry Applications and
Solving for a Specific Variable
•
•
•
•
•
•
•
•
P=a+b+c
•
180   m1  m2  m3 •
A   r2
•
C  2 r
•
A = lw
A  12 bh
Area of rectangle
Area of a triangle
Perimeter of triangle
Sum of angles of a triangle
Area of a circle
Circumference of a circle
3.5 Geometry Applications and
Solving for a Specific Variable
• Complementary
angles – add up to 90
• Supplementary angles
– add up to 180
• Vertical angles – the
angles opposite each
other are congruent
3.5 Geometry Applications and
Solving for a Specific Variable
•
1.
2.
3.
4.
Find the measure of an angle whose complement
is 10 larger.
x = degree measure of the angle.
90 – x = measure of its complement
90 – x = 10 + x
Subtract 10: 80  x  x
Add x:
80  2 x
Divide by 2: x  40
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Ratio – quotient of two quantities with the
a
same units
b
Note: percents are ratios where the second
number is always 100:
35
35%  100
 .35
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Percents :
73
73%  100
 .73
Example: If 70% of the marbles in a bag
containing 40 marbles are red, how many of the
marbles are red?:
# of red marbles =
70% of 40  (.70)  40  28
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Proportion – statement that two ratios are
equal:
a
c
b
d

Solve using cross multiplication:
ad  bc
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Solve for x:
Solution:
81
x 3

9
7
81 7  9  ( x  3)
567  9 x  27
540  9 x
x  60
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Example: d=rt (distance = rate  time)
How long will it take to drive 420 miles at 50
miles per hour?
420  50 t
t
420
50
 8.4 hours
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• General form of a mixture problem:
x units of an a% solution are mixed with
y units of a b% solution to get
z units of a c% solution
Equations will always be:
x y  z
a %( x)  b%( y )  c%( z )
3.6 Applications of Linear Equations to
Proportions, d=rt, and Mixture Problems
• Example: How many gallons of a 10% indicator solution
must be mixed with a 20% indicator solution to get 10
gallons of a 14% solution?
Let x = # gallons of 10% solution,
then 10 - x = # gallons of 20% solution :
10%( x)  20%(10  x)  14%(10)
10x  20(10  x)  14(10)
10x  200 20x  140  10x  200  140
 10x  60
x  6 gallonsof 10% m ixture
3.7 Solving Linear Inequalities in
One Variable
•
•
•
•
<

>

means “is less than”
means “is less than or equal to”
means “is greater than”
means “is greater than or equal to”
note: the symbol always points to the
smaller number
3.7 Solving Linear Inequalities in
One Variable
• A linear inequality in one variable can be
written in the form:
ax < b (a0)
• Addition property of inequality:
if a < b then a + c < b + c
3.7 Solving Linear Inequalities in
One Variable
• Multiplication property of inequality:
– If c > 0 then
a < b and ac < bc are equivalent
– If c < 0 then
a < b and ac > bc are equivalent
note: the sign of the inequality is reversed when
multiplying both sides by a negative number
3.7 Solving Linear Inequalities in
One Variable
• Example:
 x  x  x 3
2
3
1
2
1
6
 4 x  3 x  x  18
 x  x  18
 2 x  18
x  9
-9
3.8 Solving Compound Inequalities
• For any 2 sets A and B, the intersection of A and B
is defined as follows:
AB = {x  x is an element of A and x is an
element of B}
• For any 2 sets A and B, the union of A and B is
defined as follows:
AB = {x  x is an element of A or x is an element
of B}
3.8 Solving Compound Inequalities
•
Example:
3 x  1  10 and 2 x  2  4
3 x  9 and 2 x  2
x  3 and x  1
1 x  3
1
3
3.8 Solving Compound Inequalities
•
Example:
3x  1  10 or 2 x  2  4
3 x  9 or 2 x  2
x  3 or x  1
1
3