Transcript 3.1 Systems of Linear Equations 3.1 Systems of Linear Equations • Using graphs and tables to solve systems • Using substitution and elimination to.

3.1 Systems of Linear
Equations
3.1 Systems of Linear Equations
• Using graphs and tables to solve systems
• Using substitution and elimination to solve
systems
• Using systems to model data
• Value, interests, and mixture problems
• Using linear inequalities in one variable to
make predictions
Using Two Models to Make a
Prediction
• When will the life expectancy of men and
women be equal?
Years of Life
– L = W(t) = 0.114t + 77.47
– L = M(t) = 0.204t + 69.90
100
80
(84.11, 87.06)
60
40
60
80
20
Years since 1980
100
120
Equal at
approximately 87
years old in
2064.
System of Linear Equations in Two
Variables (Linear System)
• Two or more linear equations containing
two variables
y = 3x + 3
y = -x – 5
Solution of a System
• An ordered pair (a,b) is a solution of a
linear system if it satisfies both equations.
• The solution sets of a system is the set of
all solutions for that system.
• To solve a system is to find its solution set.
• The solution set can be found by finding
the intersection of the graphs of the two
equations.
Find the Ordered Pairs that Satisfy
Both Equations
• Graph both equations on the same coordinate plane
– y = 3x + 3
– y = -x – 5
• Verify
– (-3) = 3(-2) + 3
– -3 = -6 + 3
– -3 = -3
– (-3) = -(-2) – 5
– -3 = 2 – 5
– -3 = -3
Solutions for
y = 3x + 3
Solutions for
y = -x – 5
Solution for
both (-2,-3)
• Only one point satisfies
both equations
• (-2,-3) is the solution set of the system
Example
• ¾x + ⅜y = ⅞
• y = 3x – 5
• Solve first equation for y
– ¾x + ⅜y = ⅞
– 8(¾x + ⅜y) = 8(⅞)
– 24x + 24y = 56
4
8
8
– 6x -6x + 3y = 7 – 6x
– 3y = -6x + 7
3
3
3
– y = -2x + 7/3
y = 3x – 5
-.6 = 3(1.45) – 5
-.6 = 4.35 – 5
-.6 ≈ -.65
y = -2x + 7/3
-.6 = -2(1.45) + 7/3
-.6 = -2.9 + 7/3
-.6 ≈ -.57
(1.45,-.6)
Inconsistent System
• A linear system whose solution set is
empty
– Example…Parallel lines never intersect
• no ordered pairs satisfy both systems
Dependent System
• A linear system that has an infinite number
of solutions
– Example….Two equations of the same line
• All solutions satisfy both lines
y = 2x – 2
-2x + y = -2
• -2x +2x + y = -2 +2x
• y = 2x – 2
One Solution System
• There is exactly one ordered pair that
satisfies the linear system
– Example…Two lines
that intersect in only
one point
Solving Systems with Tables
x
0
1
2
3
4
y = 4x – 6
-6
-2
2
6
10
y = -6x + 14
14
8
2
-4
-10
• Since (2,2) is a solution to both equations, it is a
solution of the linear system.
3.2 Using Substitution
3.2 Using Substitution
• Isolate a variable on one side of either
equation
• Substitute the expression for the variable
into the other equation
• Solve the second equation
• Substitute the solution into one of the
equations
Example 1
1.
2.
•
•
•
•
•
y=x–1
3x + 2y = 13
3x + 2(x – 1) = 13
3x + 2x – 2 = 13
5x – 2 +2 = 13 +2
5x = 15
5
5
x=3
• y=3–1
• y=2
Solution set for the
linear system is (3,2).
Example 2
1.
2.
•
•
2x – 6y = 4
3x – 7y = 8
2x – 6y +6y = 4 +6y
2x = 6y + 4
2
•
2
2
x = 3y + 2
The solution set is (5, 1).
•
•
•
•
3(3y + 2) – 7y = 8
9y + 6 – 7y = 8
2y + 6 -6 = 8 -6
2y = 2
2
2
• y=1
• x = 3(1) + 2
• x=5
Using Elimination
• Adding left and right sides of equations
• If a=b and c=d, then a + c = b + d
– Substitute a for b and c for d
• a+c=a+c
• both sides are the same
Using Elimination
• Multiply both equations by a number so
that the coefficients of one variable are
equal in absolute value and opposite sign.
• Add the left and right sides of the
equations to eliminate a variable.
• Solve the equation.
• Substitute the solution into one of the
equations and solve.
Example 1
5x – 6y = 9
+ 2x + 6y = 12
7x + 0 = 21
7x = 21
7
2(3) + 6y = 12
6 -6 + 6y = 12 -6
6y = 12
6
6
y=2
7
x=3
Solution set for the system is (3,2)
Example 2
1. 3x + 7y = 29
2. 6x – 12y = 32
-2(3x + 7y) = -2(29)
-6x – 14y = -58
+ 6x – 12y = 32
0 – 26y = -26
-26 -26
y=1
• 3x + 7(1) = 29
• 3x + 7 -7 = 29 -7
• 3x = 22
3
3
x = 22/3
Solution (22/3, 1)
Example 3
1. 2x + 5y = 14
2. 7x – 3y = 8
3(2x + 5y) = 3(14)
6x + 15y = 42
5(7x – 3y) = 5(8)
35x – 15y = 40
Solution (2,2)
6x + 15y = 42
+ 35x – 15y = 40
41x + 0 = 82
41
41
x=2
2(2) + 5y = 14
4 -4 + 5y = 14 -4
5y = 10
5
5
y=2
Using Elimination with Fractions
2x – 5y
-1
9( 3
9 ) = ( 3 )9
3x
–
2y
7
15(
) = ( )15
15 3
5
6(-3) – 5y = -3
-18 +18 – 5y = -3 +18
-5y = 15
-5 -5
y = -3
6x – 5y = -3
-2(6x – 5y) = (-3)-2
-12x + 10y = 6
3x – 10y = 21
+___________
-9x + 0
-9
x = -3
= 27
-9
Inconsistent Systems
•If the result of substitution or elimination of a
linear system is a false statement, then the system
is inconsistent.
• y = 3x + 3
• y = 3x + 3
-1(y) = -1(3x + 3)
• y = 3x – 2
• -y = -3x – 3
• 3x + 3 = 3x – 2
+ y = 3x – 2
3x -3x + 3 = 3x -3x – 2
0= 0–2
3 ≠ - 2 False
0 ≠ -2 False
Dependent Systems
• If the result of applying substitution or
elimination to a linear system is a true statement,
then the system is dependent.
• y -3x = 3x -3x – 4
1. y = 3x – 4
• -3(-3x + y) = (-4)-3
2. -9x + 3y = -12
• -9x + 3(3x – 4) = -12 • 9x + -3y = 12
+ -9x + 3y = -12
• -9x + 9x -12 = -12
0=0
• -12 = -12
True
• True
Solving systems in one variable
• To solve an equation A = B, in one
variable, x, where A and B are
expressions,
• Solve, graph or use a table of the system
–y=A
–y=B
• Where the x-coordinates of the solutions
of the system are the solutions of the
equation A = B
Systems in one variable
•
•
•
•
•
•
•
•
y = 4x – 3
y = -x + 2
4x – 3 = -x + 2
4x +x – 3 = -x +x + 2
5x – 3 +3 = 2 +3
5x = 5
5
5
x=1
• y = 4(1) – 3
• y=1
(1,1)
Graphing to solve equations in one
variable
• -2x + 6 = 5/4x – 3
– (3,1)
• -2x + 6 = -4
– (6,-4)
(3,1)
(6,-4)
Using Tables to Solve Equations in
One Variable
• -2x + 7 = 4x – 5
• Solution (2,3)
x
y
y
-2
11
-13
-1
9
-9
0
7
-5
1
5
-1
2
3
3
3
1
7
4
-1
11
3.3 Systems to Model Data
3.3 Systems to Model Data
• Predict when the life expectance of men
and women will be the same.
• L = W(t) = .114t + 77.47
• L = M(t) = .204t + 69.90
• .114t -.114t + 77.47 = .204t + 69.90 -.114t
• 77.47 -69.90 = .09t + 69.90 -69.90
• 7.57 = .09t
• t ≈ 84.11
Solving to Make Predictions
• In 1950, there were 4 women nursing
students at a private college and 84 men.
If the number of women nursing students
increases by 13 a year and the number of
male nursing students by 6 a year. What
year will the number of male and female
students be the same?
• A = W(t) = 13t + 4
• A = M(t) = 6t + 84
Using substitution
•
•
•
•
13t + 4 = 6t + 84
13t + 4 -4 = 6t + 84 -4
13t -6t = 6t -6t + 80
7t = 80
7 7
• t ≈ 11.43
• Equal in 1961~1962
3.4 Value, Interest and Mixture
Five Step Method
1.
2.
3.
4.
5.
Define each variable
Write a system of two equations
Solve the system
Describe the results
Check
Value Problems
• Total-Value Formula
– If n objects each have a value v, then their
total value T is give by
T=vn
– Total value equals the value of the object
times the number of objects
Value Problem Example
•
A medical store is selling rolls of cloth
bandages for $4/roll and wrist splints for
$16/each. If the store sells a
combination of 72 items in a week for a
total revenue of $552, how many of each
item were sold?
1. Define the variable
•
•
Let r be the number of bandage rolls sold
Let s be the number of splits sold
2. Write a system of two equations
•
•
•
T = 4r + 16s
Substitute Total revenue to get first equation
552 = 4r + 16s
Store sold 72 items total
r + s = 72
3. Solve the system
4r + 16s = 552
r + s = 72
•
Multiply the second equation by -4
4r + 16s = 552
-4(r + s = 72)
-4r – 4s = -288
4r + 16s = 552
+ -4r – 4s = -288
0 + 12s = 264
12
12
s = 22
r + (22) = 72
r + 22 -22 = 72 -22
r = 50
4. Describe each result
•
The store sold 50 rolls of cloth bandages
and 22 wrist splints.
5. Check
 50 + 22 = 72
 4(50) + 16(22) = 552
Value Problem
• Hospital has 1,000 rooms, 300 private
rooms and 700 shared rooms. If the
private rooms cost $100 more than the
shared rooms, what is the price for each
type of room so the total revenue when all
rooms are full is $250,000?
– private room price is $100 more than shared
p = s + 100
– total revenue is $250,000
300p + 700s = 250,000
p = s + 100
300p + 700s = 250,000
300(s + 100) + 700s = 250,000
300s + 30,000 + 700s = 250,000
1,000s + 30,000 -30,000 = 250,000 -30,000
1,000s/1,000 = 220,000/1,000
s = 220
p = (220) + 100
p = 320
Model a Value Situation
•
The Monera Hospital has 1,200 rooms.
private rooms cost $240/night and public
rooms $160/night. Let x and y be room
cost respectively. Assume all rooms are
filled.
1. Let R = f(x) be the total revenue per night.
Find the equation of f.
Total revenue
R = 240x + 160y
Total number of rooms
x + y = 1,200
y = 1,200 – x
Substitute
R = 240x + 160(1,200 – x)
R = 240x + 192,000 – 160x
R = 80x + 192,000
Equation of f
f(x) = 80x + 192,000
• Graph the equation x, 0 ≤ x ≤ 1,200
• What is the slope?
– 80
• What does this represent?
– If 1 more room is reserved as private and one
less as public, then the revenue will increase
by $80.
• What is f(800)?
f(800) = 80(800) + 192,000
= 64,000 + 192,000
= 256,000
• What does this represent?
– If 800 rooms are filled as private rooms, then
the total revenue of the night would be
$256,000.
• Find f(15,000)
– f(15,000) = 80(15,000) + 192,00
= 1,392,000
– Since there are only 1,200 rooms a model
breakdown occurs.
• To make a total profit of $200,000 how
many of each room must be filled?
200,000 = 80x + 192,000
8,000 = 80x
x = 100
y = 1,200 – 100 = 1,100
100 private rooms and 1,100 public rooms must
be booked to get a total profit of $200,000.
Interest Problems
• Principal – money deposited in an account
such as a savings account, certificate of
deposit, or mutual fund
• Interest – percentage of the principal
earned when a person invests money
• Annual Simple Interest Rate – percentage
of principal that equals the interest earned
per year.
– Invest $100 and get $6.5 yr, then the simple
interest rate is 6.5%
Interest from Investment
• If a person invests $2800 at 3.5% simple
annual interest, how much interest will
they earn in a year?
.035(2800) = 98
The person will earn $98 in interest.
Interest Problems
• A person is investing in two different
accounts, he invests twice as much in an
account at 3.9% than he does in a second
account at 7.2% annual interest. Both
have 3 year averages. How much will a
person have to invest in each account to
earn a total of $395?
• If he invests twice as much at 3.9%
x = 2y
• Since the total interest is $395,
.039x + .072y = 395
• System
x = 2y
.039x + .072y = 395
• Solve the system
.039(2y) + .072y = 395
.078y + .072y = 395
.15y = 395
y = 2633.33
x = 2(2633.33)
x = 5266.66
Modeling an Interest Problem
• A person invests $2500 in a mutual fund
with a five year annual interest of 5.5%
and a CD account at 3.25% annual
interest. Let x and y be the money
invested in each account respectively.
• Let I = f(x) be the total earned from
investing $2500 for one year
• Total interest is:
I = .055x + .0325y
• I in terms of just x:
x + y = 2500
x = 2500 – y
• Substitute
I = .055x + .0325(2500 – y)
= .055x + 81.25 - .0325y
= .0225x + 81.25
• f(x) = .0225x + 81.25
• Graph the equation between 0 ≤ x ≤ 2500
• What is the slope?
– .0225
• What does this mean?
– If one more dollar is invested at 5.5% and one
less at 3.25%, the total interest will increase
by 2.25 cents
• Use a graphing calculator to create a
table. How can this help to determine how
much to invest in each account?
• If a person doesn’t know how much they
want to risk by investing in each account,
a table could help give a clearer idea of
how much to invest in each account.
• How much should be invested in each
account to earn $124 in interest?
124 = .0225x + 81.25
42.75 = .0225x
1900 = x
y = 2500 – 1900 = 600
A person should invest $1900 at 5.5% and $600
at 3.25%.
Mixture Problems
• For an x% solution of two substances that
are mixed, x% of the solution is one
substance and (100 – x)% is the other
substance.
– If 2 oz of analgesic is diluted in 10 oz of water,
2/10 = .2 = 20% of the solution is analgesic
the remaining 8/10 = .8 = 80% is water.
Mixture Problem Example
• A pharmacist needs 20oz of 15%
expectorant cough medicine, but only has
a 12% and 24% solution. How many
ounces of the 12% should he mix with the
24% solution to make 20oz of 15%
solutions?
x + y = 20
.12x + .24y = .15(20)
System
x + y = 20
.12x + .24y = 3
Solve
y = 20 – x
.12x + .24(20 – x) = 3
.12x + 4.8 - .24x = 3
-.12x + 4.8 = 3
-.12x = -1.8
x = 15
y = 20 – 15 = 5
To make 20oz 15% solution, she would need 15 oz of
the 12% solution and 5 oz of the 24% solution.
Mixture Example
• A chemist needs 6 quarts of 20% alcohol
solution, but only has a 30% alcohol solution.
How much water should he mix with the solution
to form 6 quarts of 20% solution?
x+y=6
.30x = .20(6)
.3x = 1.2
x=4
4+y=6
y=2
Chemist needs to mix 4 quarts of 30% solution with 2
quarts of water.
3.5 Using Linear Inequalities
to Make Predictions
3.5 Using Linear Inequalities to
Make Predictions
• You are debating between two car rental
companies, one charges $48 per day and
$.15/mile. The second charges $24 per
day and $.55/mile.
C = F(d) = .15d + 48
C = S(d) = .55d + 24
• Graph the equation to see for how many
miles the first company is cheaper.
• d > 60 then the first company is cheaper
Addition Property
• If a < b, then a + c < b + c
– Similar properties hold for ≤, > and ≥.
– This holds for subtraction as well
• Subtracting is same as adding negative
5<8
5<8
5+2<8+2
5-2<8-2
7 < 10
3<6
true
true
+(-c)
+(-c)
+c
+c
a
a
b
a+c b+c
a+c b+c
a
b
Multiplication Property
• For a positive number c,
– if a < b, then ac < bc.
• For a negative number c,
– if a < b, then ac > bc.
• Similar properties hold for ≤, > and ≥.
• This holds for division as well
– dividing by a nonzero number is the same as multiplying
by its reciprocal
**When multiplying or dividing both sides of an
inequality by a negative number, reverse the
inequality symbol.
Multiplication Property (cont)
5(2) < 8(2)
10 < 16
true
4/2 < 8/2
2<4
true
5(-2) < 8(-2)
-10 < -16
false
-10 > -16
4/(-2) < 8/(-2)
-2 < -4
false
-2 > -4
Linear inequalities in One Variable
• A linear inequality in one variable is an
inequality that can be put into one of these
forms:
– mx + b < 0
– mx + b ≤ 0
– mx + b > 0
– mx + b ≥ 0
m and b are constants and m ≠ 0
• A number satisfies an inequality, if the
inequality becomes a true statement
Solutions to Inequalities
• 6x – 14 < 16
– Does 3 satisfy the equation?
6(3) – 14 < 16
18 – 14 < 16
4 < 16
true
Yes, 3 satisfies the equation
– Does 8 satisfy the equation?
6(8) – 14 < 16
48 – 14 < 16
34 < 16
false
No, 8 doesn’t satisfy the equation
Inequality Examples
• 6x – 14 < 16
– Find all solutions of the inequality
6x – 14 < 16
6x < 30
x<5
• -5x + 21 > 11
-5x /(-5) > -10/(-5)
x<2
**Remember to change the sign when dividing
or multiplying by a negative number.
Inequalities on timelines
• x<2
• x≤2
• x>2
(-∞,2)
0
2
x
(-∞,2]
0
0
• x≥2
2
x
(2,∞)
2
– ( or ) represents < or >
– [ or ] represents ≤ or ≥
x
[2,∞)
0
2
• Interval – set of real
numbers represented
by the number line or
by an unbroken portion
of it
• Interval notation
x
– All real numbers is
represented by
(- ∞ ,∞)
Solving a Linear Inequality
-2(3x – 6) < 20 – 2x
-6x + 12 -12 < 20 – 2x -12
-6x +2x < 8 – 2x +2x
-4x /(-4) < 8 /(-4)
x>2
0
2
(2,∞)
x
Three-part Inequalities
• 2<x<4
0
2
(2,4)
4
• 2≤x≤4
0
2
[2,4]
4
• 2<x≤4
0
2
2
x
(2,4]
4
• 2≤x<4
0
x
x
[2,4)
4
x
Solving Three-part Inequalities
•
•
•
•
-2 ≤ -6x – 8 < 16
-2 +8 ≤ -6x – 8 +8 < 16 +8
6 /(-6) ≤ -6x /(-6) < 24 /(-6)
-1 ≥ x > -4
-4
• (-4,-1]
-2
0
x
Using Inequalities to Model
Comparisons between two quantities
• You are debating between two car rental
companies, one charges $48 per day and
$.15/mile. The second charges $24 per
day and $.55/mile.
– C = F(d) = .2d + 48
– C = S(d) = .55d + 24
• For how many miles the first company is
cheaper?
– F(d) < S(d)
.15d + 48 < .55d + 24
.15d + 48 -24 < .55d + 24 -24
.15d -.15d + 24 < .55d -.15d
24 < .40d
60 < d
The first company offers the lower price if
the car is driven over 60 miles. (Same as
original example)