Systems of Linear Equations and Inequalities

1
Systems of Linear Equations and Inequalities

• Chapter 4

2
Chapter Sections
4.1 Solving Systems of Linear Equations in Two
Variables by Graphing 4.2 Solving Systems of
Linear Equations in Two Variables Using
Substitution 4.3 Solving Systems of Linear
Equations in Two Variables Using Elimination 4.4
Solving Direct Translation, Geometry, and
Uniform Motion Problems Using Systems of Linear
Equations in Two Variables 4.5 Solving Mixture
Problems Using Systems of Linear Equations in Two
Variables 4.6 Systems of Linear Inequalities
3
Solving Systems of Linear Equations in Two
Variables by Graphing

• Section 4.1

4
Solutions
A system of linear equations is a grouping of two
or more linear equations where each equation
contains one or more variables.
A brace is used to remind us that we are dealing
with a system of equations.
A solution of a system of equations consists of
values for the variables that satisfy each
equation of the system.
5
Solutions
Example Determine whether ( 4, 16) is a
solution to the system of equations.
y 4x 16 4( 4) 16 16
y 2x 8 16 2( 4) 8 16 8 8 16
16
Yes, it is a solution.
?
?
6
Solving a System by Graphing
Example Solve the following system by graphing.
Continued.
7
Solving a System by Graphing
Example continued
y 2x ? 1
Check
y ?3x ? 6
y ? 3x ? 6
?3 ? 3(?1) ? 6
?3 3 ? 6
?3 ?3
?
y 2x ? 1
?3 2(?1) ? 1
The point, or points, at which two lines
intersect, if any, represents the solution of the
system of equations.
?3 ?2 ? 1
?3 ?3
?
8
Solving a System by Graphing
Steps for Obtaining the Solution of a System
of Linear Equations by Graphing Step 1 Graph
the first equation in the system. Step 2 Graph
the second equation in the system. Step 3
Determine the point of intersection, if any. Step
4 Verify that the point of intersection
determined in Step 3 is a solution of the system.
Remember to check the point in both equations.
9
Inconsistent Systems
An inconsistent system is a system of equations
that has no solution.
3x ? 2y ?4
? 9x 6y ?1
The system has no solution, so the solution set
is or ?.
10
Dependent Systems
A dependent system is a system of equations where
the solution set is the set of all points on the
line.
4x ? 6y 8
? 2x 3y ?4
There are infinitely many solutions to this
system.
11
Classifying Systems Graphically
Consistent The equations are independent.
If the lines intersect, the system of equations
has one solution given by the point of
intersection.
(3, 5)
Two lines intersect at one point.
Inconsistent
If the lines are parallel, then the system of
equations has no solution because the lines never
intersect.
Parallel lines
Consistent The equations are dependent.
If the lines lie on top of each other, then the
system has infinitely many solutions. The
solution set is the set of all points on the line.
Lines coincide
12
Classifying Systems Algebraically
Example Without graphing, determine the number
of solutions of the system.
Find the slope and y-intercept of each equation
to determine the number of solutions.
y ?x 5
y x ? 5
The slope of each line is different, so the lines
will intersect at one point and there will be one
solution.
13
Classifying Systems Algebraically
Example Without graphing, determine the number
of solutions of the system.
5x y ? 1 10x 2y ? 2
Rewrite each equation in slope-intercept form to
determine the slope and y-intercept of each
equation.
5x y ? 1
10x 2y ? 2
y ?5x ? 1
y ?5x ? 1
The slope and the y-intercept of each line is the
same so the lines are the same and there is an
infinite number of solutions.
14
Solving Applications
Example Amtech has two different cellular phone
plans. Plan A charges a monthly fee of 14.95
plus 0.05 per minute. Plan B charges a monthly
fee of 18.95 and 0.03 per minute. Determine
the number of minutes for which the cost of the
plans will be the same.
Step 1 Identify We want to know the number of
minutes that can be used such that the two plans
have the same fee.
Step 2 Name Let m represent the number of
minutes used and let C represent the monthly cost.
Continued.
15
Solving Applications
Example continued
Step 3 Translate
Monthly cost monthly fee per minute rate
of minutes
Plan A C 14.95 0.05m
Plan B C 18.95 0.03m
Step 4 Solve Create a table which shows the
monthly cost for various minutes used for each
plan.
Continued.
16
Solving Applications
Example continued
Graph the system of equations.
(300, 29.95)
(150, 23.45)
Plan B
(300, 27.95)
(150, 22.45)
Plan A
(200, 24.95)
The lines intersect at (200, 24.95).
Continued.
17
Solving Applications
Example continued
Step 5 Check
Plan A C 14.95 0.05m
When 200 minutes are used C 14.95
0.05(200)
24.95
Plan B C 18.95 0.03m
When 200 minutes are used C 18.95
0.03(200)
24.95
?
Step 6 Answer The cost of the plans will be the
same when 200 minutes are used.
18
Solving Systems of Linear Equations in Two
Variables Using Substitution

• Section 4.2

19
The Substitution Method
Steps for Solving a System of Two Linear
Equations Containing Two Unknowns by Substitution
Step 1 Solve one of the equations for one of the
unknowns.
Step 2 Substitute the expression solved for in
Step 1 into the other equation. The result will
be a single linear equation in one unknown.
Step 3 Solve the linear equation in one unknown
found in Step 2.
Step 4 Use one of the original equations to find
the value of the other variable.
Step 5 Check your answer.
20
The Substitution Method
Example Solve the following system by
substitution
y 2x 13
Solve the first equation for y.
4x 9y 7
Equation (2)
4x 9(2x 13) 7
Substitute 2x 13 into equation (2).
4x 18x 117 7
Distribute.
22x 110
Combine like terms.
x 5
Substitute x 5 into one of the original
equations.
Continued.
21
The Substitution Method
Example continued
2x y 13
Equation (1).
Substitute x 5 into equation (1).
2(5) y 13
Simplify.
10 y 13
y 3
Subtract 10 from both sides.
y 3
Divide both sides by 1.
Check the solution (5, 3) in both original
equations.
4x 9y 7
2x y 13
2(5) (?3) 13
4(5) 9(?3) 7
10 3 13
20 27 7
13 13
7 7
?
?
22
The Substitution Method
(1)
Example Solve the following system by
substitution
(2)
Multiply each term in equation (2) by the LCD.
Simplify.
Solve equation (2) for y.
Substitute 2x 9 into equation (1).
Simplify.
Continued.
23
The Substitution Method
Example continued
Combine like terms and add 18 to each side.
Simplify.
Equation (1)
Substitute x 4 into equation (1).
Simplify.
Check the solution (?4, ?1) in both original
equations.
Continued.
24
The Substitution Method
Example continued
Check the solution (4, 1) in both original
equations.
?
?
25
Inconsistent Systems
Example Solve the following system by
substitution
3x 6y 12 (1) x 2y 7 (2)
x ?2y 7
Solve equation (2) for x.
3(?2y 7) 6y 12
Substitute 2y 7 into equation (1).
Simplify.
?6y 21 6y 12
Simplify.
21 12
The system has no solution, so the solution set
is or ?.
The system is inconsistent.
26
Dependent Systems
Example Solve the following system by
substitution
x y 1 (1) 3x 3y 3 (2)
Solve equation (1) for y.
y ?x 1
3x 3( x 1) 3
Substitute x 1 into equation (2).
Distribute.
3x 3x 3 3
Simplify.
3 3
The system has infinitely many solutions.
The system is consistent, but dependent.
27
Solving Applications
Example Wilsons Woodworks produces desk chairs
that they sell to local home improvement stores.
The cost y (in hundreds of dollars) to produce x
desk chairs can be determined by the equation y
0.5x 6. The revenue y (in hundreds of dollars)
from the sale of x desk chairs is given by the
equation y 1.5x. Find the break-even point.
Step 1 Identify We want to know the number of
desk chairs that need to be sold so that the
company can break even. (The break-even point is
the number of units of a product that is produced
so that the cost of production will be equal to
the amount of revenue from the sale of the
product.)
Step 2 Name Let x represent the number of desk
chairs. Let y represent the cost/revenue.
Continued.
28
Solving Applications
Example continued
Step 3 Translate The two equations are given
Step 4 Solve Use substitution to solve the
system of equations.
y 0.5x 6
Equation (1)
1.5x 0.5x 6
Substitute 1.5x into equation (1).
x 6
Solve for x.
Continued.
29
Solving Applications
Example continued
Step 5 Check
Substitute x 6 into each original equation.
y 0.5x 6
y 1.5x
0.5(6) 6
1.5(6)
9
9
?
Step 6 Answer Wilsons Woodworks will need to
produce and sell 6 desk chairs to break even.
30
Solving Systems of Linear Equations in Two
Variables Using Elimination

• Section 4.3

31
The Elimination Method
Steps for Solving a System of Linear Equations by
Elimination
Step 1 Make sure that the coefficients on one of
the variables are additive inverses by
multiplying (or dividing) both sides of one (or
both) equation(s) by a nonzero constant.
Step 2 Add the equations to eliminate the
variable whose coefficients are now additive
inverses. Solve the resulting equation for the
remaining unknown.
Step 3 Substitute the value of the variable
found in Step 2 into one of the original
equations to find the value of the remaining
variable.
Step 4 Check your answer.
32
The Elimination Method
Example Solve the following system by
elimination
(1) (2)
5x 3y 14 2x y 6
Multiply both sides of equation (2) by ?3.
?3(2x y) ?3(6)
Simplify.
? 6x 3y ?18
5x 3y 14 ? 6x 3y ?18
Add equations (1) and (2).
? x ? 4
x 4
Divide both sides by 1.
Substitute 4 into equation (2).
2(4) y 6
Solve for y.
y 2
Continued.
33
The Elimination Method
Example continued
Check the solution (4, 2) in both original
equations.
5x 3y 14
2x y 6
5(4) 3(2) 14
2(4) 2 6
20 6 14
8 2 6
14 14
6 6
?
?
34
The Elimination Method
Example Solve the following system by
elimination
Multiply both sides of equation (1) by 8.
Simplify.
Multiply both sides of equation (1) by 3.
Simplify.
Equations (1) and (2) now contain additive
inverses.
Continued.
35
The Elimination Method
Example continued
Add equation (1) and (2).
Divide both sides by 52.
Substitute x ?1 into the original equation (2).
Add 16 to both sides.
Divide both sides by 3.
Continued.
36
The Elimination Method
Example continued
Check the solution ( 1, 4) in both original
equations.
?
?
37
Inconsistent Systems
Example Solve the following system by
elimination
5x y 3 (1) 10x 2y 2 (2)
Multiply equation (1) by 2.
2(5x y) 2(3)
Simplify.
10x 2y 6
10x 2y 6
Add equation (1) and (2).
10x 2y 2
0 8
The system has no solution, so the solution set
is or ?.
The system is inconsistent.
38
Dependent Systems
Example Solve the following system by
elimination
Multiply equation (1) by 6.
6(6x 4y) 6(8)
Simplify.
36x 24y 48
Multiply equation (2) by 4.
4(9x 6y) 4(12)
Simplify.
36x 24y 48
36x 24y 48
36x 24y 48
Add equation (1) and (2).
0 0
The system has infinitely many solutions.
The system is consistent, but dependent.
39
Solving Applications
Example Findlay Community Center is holding
their annual musical. 650 tickets were sold for
a total value of 4375. If adult tickets cost
7.50, and student tickets cost 3.50, how many
of each kind of ticket were sold?
Step 1 Identify We want to know the number of
adult tickets and the number of student tickets
sold for the musical.
Step 2 Name Let x represent the number of
adult tickets. Let y represent the number of
student tickets.
Continued.
40
Solving Applications
Example continued
Step 3 Translate There were 650 total tickets
sold.
x y 650
Each adult ticket costs 7.50 and each student
ticket costs 3.50. The total value of the sold
tickets was 4375.
7.5x 3.5y 4375
Step 4 Solve Use addition to solve the system
of equations.
7.5(x y) (650) 7.5
Multiply equation (1) by 7.5.
7.5x 7.5y 4875
Simplify.
Continued.
41
Solving Applications
Example continued
7.5x 7.5y 4875
(1) (2)
7.5x 3.5y 4375
Add equation (1) and (2).
4y 500
Divide both sides by 4.
y 125
Substitute 125 into the original equation (1).
x 125 650
Simplify.
x 525
Continued.
42
Solving Applications
Example continued
Step 5 Check Substitute x 525 and y 125
into the original equations.
x y 650
7.5x 3.5y 4375
525 125 650
7.5(525) 3.5(125) 4375
650 650
4375 4375
?
?
Step 6 Answer 525 adult tickets and 125 student
tickets were sold.
43
Solving Direct Translation, Geometry, and Uniform
Motion Problems Using Systems of Linear Equations
in Two Variables

• Section 4.4

44
Direct Translation Problems

• Example
• Find two numbers whose sum is 56 and when three
  times the first is subtracted from the second,
  the difference is 4.

Step 1 Identify We are looking for two unknown
numbers.
Step 2 Name
Let x the first number.
Let y the second number.
Step 3 Translate The sum of the two numbers is
56.
x y 56
The second number minus three times the first
number is 4.
y 3x 4
Continued.
45
Direct Translation Problems
Example continued
The system of equations is
Step 4 Solve Using the substitution method will
work easily since the coefficient of y is 1 in
equation (2).
y 3x 4
Equation (2)
Solve equation (2) for y.
y 3x 4
Equation (1)
x y 56
Substitute 3x 4 into equation (1).
x 3x 4 56
Continued.
46
Direct Translation Problems
Example continued
x 3x 4 56
Simplify.
4x 4 56
Subtract 4 from each side.
4x 52
Divide each side by 4.
x 13
Substitute x 13 into equation (1).
13 y 56
Subtract 13 from both sides.
y 43
Continued.
47
Direct Translation Problems
Example continued
x 13
y 43
Step 5 Check
x y 56
y 3x 4
43 3(13) 4
13 43 56
43 39 4
56 56
?
4 4
?
Step 6 Answer The two numbers are 13 and 43.

48
Geometry Problems

• Example
• The perimeter of a rectangular field is 142 feet.
  If the length of the field is 15 feet longer
  than the width, find the dimensions of the field.

Step 1 Identify We want to know the length and
the width of the field.
Step 2 Name
Let w the width of the field.
Let l the length of the field.
Continued.
49
Geometry Problems
Example continued
Step 3 Translate Use the formula for the
perimeter of a rectangle.
P 2l 2w
142 2l 2w
The length of the field is 15 feet longer than
the width.
l w 15
The system of equations is
Continued.
50
Geometry Problems
Example continued
Step 4 Solve The substitution method will work
the best.
Substitute w 15 into (1).
2(w 15) 2w 142
Distribute.
2w 30 2w 142
Simplify.
4w 112
Divide each side by 4.
w 28
l 28 15
Substitute w 28 into (2).
Simplify.
l 43
Continued.
51
Geometry Problems
Example continued
w 28
l 43
Step 5 Check
2l 2w 142
l w 15
43 28 15
2(43) 2(28) 142
43 43
86 56 142
?
142 142
?
Step 6 Answer The width of the field is 28
feet and the length is 43 feet.

52
Uniform Motion Problems
Example Kathy drove her car from Salem to
Granger, a distance of 360 miles. Terry drove
his car from Salem to Dartsville, a distance of
280 miles in the same time. Kathy drove 20 miles
per hour faster than Terry on her trip. What was
the average speed in miles per hour for each
driver?
Step 1 Identify
Distance problems can be solved using the
formula distance rate time (d rt).
Step 2 Name
Let r the rate (speed) of Terrys car.
Let r 20 the rate (speed) of Kathys car
since Kathy drove 20 mph faster.
The time, t, for each driver was the same.
Continued.
53
Uniform Motion Problems
Example continued
Step 3 Translate
Step 4 Solve
Since the time for each driver was the same, we
can set the times equal to each other.
Continued.
54
Uniform Motion Problems
Example continued
Multiply both sides by r(r 20).
Simplify.
Distribute.
Subtract 280r from each side.
Divide each side by 80.
Continued.
55
Uniform Motion Problems
Example continued
Step 5 Check
Terrys distance was 280 miles and her speed was
70 mph.
Kathys distance was 360 miles and her speed was
70 20 90 mph.
?
Step 6 Answer
Terrys speed was 70 mph. Kathys speed was 90
mph.
56
Solving Mixture Problems Using Systems of Linear
Equations in Two Variables

• Section 4.5

57
Mixture Problems
Problems that involve mixing two or more
substances are called mixture problems.
This type of problem can be set up using the model
number of units of the same kind rate amount
A table is helpful for organizing the given
information.
58
Mixture Problems

• Example
• Matthew received an 8000 Christmas bonus from
  his employer. He invested part of the bonus in a
  savings account earning 3 simple interest per
  year and the rest in a certificate of deposit
  (CD) that earns 4.5 simple interest annually.
  At the end of the year, he will receive 330
  interest on his investments. How much money did
  Matthew invest in each account?

Step 1 Identify We want to know how much money
was put into the saving account at 3 interest
and how much was put into the CD earning 4.5
interest.
Step 2 Name
Let x the amount of money deposited into the
savings account.
Let y the amount of money deposited into the CD.
Continued.
59
Mixture Problems
Example continued
Step 3 Translate Summarize the information in a
table.
Total amount invested
x y 8000
The interest earned at 3 is 0.03x. The
interest earned at 4.5 is 0.045y.
Total interest earned
0.03x 0.045y 330
Continued.
60
Mixture Problems
Example continued
The system of equations is
(1) (2)
x y 8000
0.03x 0.045y 330
Step 4 Solve The elimination method will work
easily.
Multiply equation (1) by 0.03.
0.03( x y 8000)
Simplify.
0.03x 0.03y 240
Add equation (1) and (2).
0.03x 0.045y 330
Simplify.
0.015y 90
Divide both sides by 0.015.
y 6000
Continued.
61
Mixture Problems
Example continued
y 6000
x 6000 8000
Substitute 6000 into equation (1).
x 2000
Subtract 6000 from both sides.
Step 5 Check
x y 8000
0.03x 0.045y 330
2000 6000 8000
0.03(2000) 0.045(6000) 330
8000 8000
60 270 330
?
330 330
?
Step 6 Answer 2000 was invested in the savings
account earning 3 simple interest per year and
6000 was invested in the CD earning 4.5 simple
interest annually.
62
Mixture Problems

• Example
• A nut company wants to produce 20 pounds of nuts
  worth 4.50 per pound. To obtain this, they want
  to mix nuts worth 5.00 per pound with nuts worth
  3.00 per pound. How many pounds of each type
  should be used to make the desired 20 pounds
  worth 4.50 per pound?

Step 1 Identify We want to know how many pounds
of each type of nuts to use to obtain 20 pounds
of nuts worth 4.50 per pound.
Step 2 Name
Let x the number of pounds of nuts worth 5.00
per pound.
Let y the number of pounds of nuts worth 3.00
per pound.
Continued.
63
Mixture Problems
Example continued
Step 3 Translate Summarize the information in a
table.
Total pounds of nuts
x y 20
5x 3y 4.5(20)
Total value of the 20 pounds
Continued.
64
Mixture Problems
Example continued
The system of equations is
x y 20 (1)
5x 3y 90 (2)
Step 4 Solve
Solve equation (1) for x.
x 20 y
Substitute 20 y into (2).
5(20 y) 3y 90
Distribute.
100 5y 3y 90
Simplify.
2y 10
Simplify.
y 5
Substitute y 5 into (1).
x 5 20
Simplify.
x 15
Continued.
65
Mixture Problems
Example continued
x 15
y 5
Step 5 Check
x y 20
5x 3y 90
15 5 20
5(15) 3(5) 90
20 20
75 15 90
?
90 90
?
Step 6 Answer 15 pounds of the 5.00 per pound
nuts and 5 pounds of the 3.00 per pound nuts
should be used to obtain 20 pounds of nuts worth
4.50 per pound.
66
Systems of Linear Inequalities

• Section 4.6

67
Solutions
An ordered pair satisfies a system of linear
inequalities if it makes each inequality in the
system a true statement.

• Example
• Determine if the ordered pair (3, 2) is a
  solution to the system of linear inequalities.

(3, 2) is a solution to the system.
True
True
68
Graphing a System of Linear Inequalities
Example Graph the solution of the system
The solution will be the set of all points that
satisfy both of the inequalities in the system.
69
Graphing a System of Linear Inequalities
Example Graph the solution of the system.
3x y 9
3x 5y 15
70
Graphing a System of Linear Inequalities
Example Graph the solution of the system
y gt 2x 4
y lt 2x 2
y 2x 4
y 2x 2
71
Applications
Example A certain breed of dog needs at least 10
grams of protein and 8 grams of fat per day.
Alfalfa Dog Morsels provides 6 grams of protein
and 2 grams of fat per serving, while Chunky
Moist Meats provides 2 grams of protein and 3
grams of fat per serving. Write a system of
linear inequalities that represents the possible
combination of the two types of food.
Step 1 Identify We want to know the possible
combinations of the two types of food that will
yield the required grams of protein and fat.
Continued.
72
Applications
Step 2 Name
Let x the number of servings of Alfalfa Dog
Morsels.
Let y the number of servings of Chunky Moist
Meats.
Step 3 Translate If Alfalfa Dog Morsels is
used, the amount of protein is 6x. If Chunky
Moist Meats is used, the amount of protein is 2y.
The total amount of protein required is at
least 10 grams.
6x 2y ? 10
Similarly, if Alfalfa Dog Morsels is used, the
amount of fat is 2x. If Chunky Moist Meats is
used, the amount of fat is 3y. The total amount
of fat required is at least 8 grams.
2x 3y ? 8
Continued.
73
Applications
Example continued
The number of servings of each type of food can
never be negative so we have the nonnegativity
constraints.
x ? 0
y ? 0
We now have the following system of linear
inequalities
Continued.
74
Applications
Example continued
Step 4 Solve
6x 2y 10
2x 3y 8
The following solution shows the possible
combinations of servings of the two types of dog
food.
Continued.
75
Applications
Example continued
Step 5 Check Choose a point in the solution and
determine if all of the inequalities in the
system are satisfied.
2x 3y ? 8
6x 2y ? 10
6(4) 2(4) ? 10
2(4) 3(4) ? 8
24 8 ? 10
8 12 ? 8
32 ? 10
20 ? 8
?
?
x ? 0
y ? 0
4 ? 0
4 ? 0
?
?