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Lesson 3-1 Solving Systems of Equations
by Graphing
Lesson 3-2 Solving Systems of Equations
Algebraically
Lesson 3-3 Solving Systems of Inequalities
by Graphing
Lesson 3-4 Linear Programming
Lesson 3-5 Solving Systems of Equations in
Three Variables
Five-Minute Check (over Chapter 2)
Main Ideas and Vocabulary
California Standards
Example 1: Solve the System of Equations by
Completing a Table
Example 2: Solve by Graphing
Example 3: Break-Even Point Analysis
Example 4: Intersecting Lines
Example 5: Same Line
Example 6: Parallel Lines
Concept Summary
• Solve systems of linear equations by graphing.
• Determine whether a system of linear equations is
consistent and independent, consistent and
dependent, or inconsistent.
• system of equations
• consistent
• inconsistent
• independent
• dependent
Standard 2.0 Students solve systems of linear
equations and inequalities (in two or three variables)
by substitution, with graphs, or with matrices. (Key)
Solve the System of Equations
by Completing a Table
Solve the system of equations by completing
a table.
x+y=3
–2x + y = –6
Solve for y in each equation.
x+y = 3
y = –x + 3
–2x + y = –6
y = 2x – 6
Solve the System of Equations
by Completing a Table
Use a table to find the solution that satisfies both
equations.
Answer: The solution to the system is (3, 0).
What is the solution of the system of equations?
x+y=2
x – 3y = –6
B
C. (2, 0)
C.
C
D. (–4, 6)
D.
D
0%
0%
0%
0%
D
B.
C
B. (0, 2)
B
A. A
A
A. (1, 1)
Solve by Graphing
Solve the system of equations by graphing.
x – 2y = 0
x+y=6
Write each equation in slope-intercept form.
The graphs appear to
intersect at (4, 2).
Solve by Graphing
Check Substitute the coordinates into each equation.
x – 2y = 0
?
x+y =6
?
4 – 2(2) = 0
4+2 =6
0=0
6=6
Original equations
Replace x with 4
and y with 2.
Simplify.
Answer: The solution of the system is (4, 2).
Which graph shows the solution to the system of
equations below?
x + 3y = 7
x–y = 3
A. A
A.
C.
B. B
C. C
D. D
0%
0%
D
0%
C
0%
B
D.
A
B.
Break-Even Point Analysis
SALES A service club is selling copies of its holiday
cookbook to raise funds for a project. The printer’s
set-up charge is $200, and each book costs $2 to print.
The cookbooks will sell for $6 each. How many must
the club sell before it makes a profit?
Let x = the number of cookbooks, and let
y = the number of dollars.
Cost of books
is
cost per book
plus
set-up charge.
y
=
2x
+
200
Break-Even Point Analysis
Income from
books
y
is
price per
book
times
number of
books.
=
6

x
Answer:
The graphs intersect at
(50, 300). This is the
break-even point. If the
group sells fewer than 50
books, they will lose
money. If the group sells
more than 50 books, they
will make a profit.
The student government is selling candy bars. It cost
$1 for each candy bar plus a $60 set-up fee. The
group will sell the candy bars for $2.50 each. How
many do they need to sell to break even?
B.
B
C. 60
C.
C
D. 80
D.
D
A
0%
0%
0%
0%
D
B. 40
C
A. A
B
A. 0
Intersecting Lines
Graph the system of equations and describe it as
consistent and independent, consistent and
dependent, or inconsistent.
x–y=5
x + 2y = –4
Write each equation in slope-intercept form.
Intersecting Lines
Answer:
The graphs of the equations intersect at (2, –3). Since
there is one solution to this system, this system is
consistent and independent.
D
C
B
A
Graph the system of equations below. What type of
system of equations is shown?
x+y=5
2x = y – 11
A. A
A. consistent and independent
B. B
B. consistent and dependent
C. C
C. consistent
0%
0%
0%
0%
D. D
D. none of the above
Same Line
Graph the system of equations and describe it as
consistent and independent, consistent and
dependent, or inconsistent.
9x – 6y = –6
6x – 4y = –4
Write each equation in slope-intercept form.
Since the equations are equivalent, their graphs
are the same line.
Same Line
Answer:
Any ordered pair representing a point on that line will
satisfy both equations. So, there are infinitely many
solutions. This system is consistent and dependent.
D
C
B
A
Graph the system of equations below. What type of
system of equations is shown?
x+y=3
2x = –2y + 6
A. A
A. consistent and independent
B. B
B. consistent and dependent
C. C
C. inconsistent
0%
0%
0%
0%
D. D
D. none of the above
Parallel Lines
Graph the system of equations and describe it as
consistent and independent, consistent and
dependent, or inconsistent.
15x – 6y = 0
5x – 2y = 10
Write each equation in slope-intercept form.
Parallel Lines
Answer:
The lines do not intersect. Their graphs are parallel lines.
So, there are no solutions that satisfy both equations.
This system is inconsistent.
D
C
B
A
Graph the system of equations below. What type of
system of equations is shown?
y = 3x + 2
–6x + 2y = 10
A. A
A. consistent and independent
B. B
B. consistent and dependent
C. C
C. inconsistent
0%
0%
0%
0%
D. D
D. none of the above
Five-Minute Check (over Lesson 3-1)
Main Ideas and Vocabulary
California Standards
Example 1: Solve by Using Substitution
Example 2: Standards Example: Solve by
Substitution
Example 3: Solve by Using Elimination
Example 4: Multiply, Then Use Elimination
Example 5: Inconsistent System
• Solve systems of linear equations by using
substitution.
• Solve systems of linear equations by using
elimination.
• substitution method
• elimination method
Standard 2.0 Students solve systems of linear
equations and inequalities (in two or three variables)
by substitution, with graphs, or with matrices. (Key)
Solve by Using Substitution
Use substitution to solve the system of equations.
x + 4y = 26
x – 5y = –10
Solve the first equation for x in terms of y.
x + 4y = 26
x = 26 – 4y
First equation
Subtract 4y from each side.
Solve by Using Substitution
Substitute 26 – 4y for x in the second equation and
solve for y.
x – 5y = –10
26 – 4y – 5y = –10
–9y = –36
y=4
Second equation
Substitute 26 – 4y for x.
Subtract 26 from each side.
Divide each side by –9.
Solve by Using Substitution
Now substitute the value for y in either of the original
equations and solve for x.
x + 4y = 26
x + 4(4) = 26
x + 16 = 26
x = 10
First equation
Replace y with 4.
Simplify.
Subtract 16 from each side.
Answer: The solution of the system is (10, 4).
Solve the system of equations using substitution.
What is the solution to the system of equations?
x – 3y = 2
x + 7y = 12
D. (5, 1)
C.
C
D.
D
0%
0%
0%
0%
D
B
C
C. (8, 2)
B.
B
B.
A. A
A
A. (1, 5)
Solve by Substitution
Lancaster Woodworkers Furniture Store builds two
types of wooden outdoor chairs. A rocking chair
sells for $265 and an Adirondack chair with
footstool sells for $320. The books show that last
month, the business earned $13,930 for the 48
outdoor chairs sold. How many of each chair were
sold?
Read the Item
You are asked to find the number of each type of
chair sold.
Solve by Substitution
Solve the Item
Step 1 Define variables and write the system of
equations. Let x represent the number of
rocking chairs sold and y represent the number
of Adirondack chairs sold.
x + y = 48
265x + 320y = 13,930
The total number of chairs
sold was 48.
The total amount earned
was $13,930.
Solve by Substitution
Step 2 Solve one of the equations for one of the
variables in terms of the other. Since the
coefficient of x is 1, solve the first equation for
x in terms of y.
x + y = 48
x = 48 – y
First equation
Subtract y from each side.
Solve by Substitution
Step 3 Substitute 48 – y for x in the second equation.
265x + 320y = 13,930
265(48 – y) + 320y = 13,930
12,720 – 265y + 320y = 13,930
55y = 1210
y = 22
Second equation
Substitute 48 – y for x.
Distributive Property
Simplify.
Divide each side by 55.
Solve by Substitution
Step 4 Now find the value of x. Substitute the value
for y into either equation.
x + y = 48
x + 22 = 48
x = 26
First equation
Replace y with 22.
Subtract 22 from each side.
Answer: They sold 26 rocking chairs and 22
Adirondack chairs.
D
C
B
A
AMUSEMENT PARKS At Amy’s Amusement Park,
tickets sell for $24.50 for adults and $16.50 for
children. On Sunday, the amusement park made
$6405 from selling 330 tickets. How many of each
kind of ticket was sold?
A. A
A. 210 adult; 120 children
B. B
B. 120 adult; 210 children
C. C
C. 300 children; 30 adult
0%
0%
0%
0%
D. D
D. 300 children; 30 adult
Solve by Using Elimination
Use the elimination method to solve the system
of equations.
x + 2y = 10
x+y=6
In each equation, the coefficient of x is 1. If one equation
is subtracted from the other, the variable x will be
eliminated.
x + 2y = 10
(–)x + y = 6
y= 4
Subtract the equations.
Solve by Using Elimination
Now find x by substituting 4 for y in either original
equation.
x+y =6
Second equation
x+4 =6
Replace y with 4.
x =2
Subtract 4 from each side.
Answer: The solution is (2, 4).
Use the elimination method to solve the system
of equations. What is the solution to the system?
x + 3y = 5
x + 5y = –3
B
C. (2, 1)
C.
C
D. no solution
D.
D
0%
0%
0%
0%
D
B.
C
B. (17, –4)
B
A. A
A
A. (2, –1)
Multiply, Then Use Elimination
Use the elimination method to solve the system of
equations.
2x + 3y = 12
5x – 2y = 11
Multiply the first equation by 2 and the second equation
by 3. Then add the equations to eliminate the y variable.
2x + 3y = 12
Multiply by 2.
5x – 2y = 11
Multiply by 3.
4x + 6y = 24
(+)15x – 6y = 33
19x
= 57
x=3
Multiply, Then Use Elimination
Replace x with 3 and solve for y.
2x + 3y = 12
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y=2
First equation
Replace x with 3.
Multiply.
Subtract 6 from each side.
Divide each side by 3.
Answer: The solution is (3, 2).
Use the elimination method to solve the system
of equations. What is the solution to the system of
equations?
x + 3y = 7
2x + 5y = 10
A.
A. A
B. (1, 2)
B.
B
D.
D
0%
D
D. no solution
0%
C
C
0%
B
C.
A
C. (–5, 4)
0%
Inconsistent System
Use the elimination method to solve the system of
equations.
–3x + 5y = 12
6x – 10y = –21
Use multiplication to eliminate x.
–3x + 5y = 12
6x – 10y = –21
Multiply by 2.
–6x + 10y = 24
(+)6x – 10y = –21
0= 3
Answer: Since there are no values of x and y that will
make the equation 0 = 3 true, there are no
solutions for the system of equations.
Use the elimination method to solve the system
of equations. What is the solution to the system of
equations?
2x + 3y = 11
–4x – 6y = 20
A. (1, 3)
A. A
B. (–5, 0)
B.
B
D. no solution
D.
D
0%
D
C
0%
C
C.
A
C. (2, –2)
0%
B
0%
Five-Minute Check (over Lesson 3-2)
Main Ideas and Vocabulary
California Standards
Example 1: Intersecting Regions
Example 2: Separate Regions
Example 3: Write and Use a System of Inequalities
Example 4: Find Vertices
• Solve systems of inequalities by graphing.
• Determine the coordinates of the vertices of a
region formed by the graph of a system of
inequalities.
• system of inequalities
Standard 2.0 Students solve systems of linear
equations and inequalities (in two or three variables)
by substitution, with graphs, or with matrices. (Key)
Intersecting Regions
A. Solve the system of inequalities by graphing.
y ≥ 2x – 3
y < –x + 2
Solution of y ≥ 2x – 3 → Regions 1 and 2
Solution of y < –x + 2 → Regions 2 and 3
Intersecting Regions
A.
Answer: The intersection of these regions is Region 2,
which is the solution of the system of
inequalities. Notice that the solution is a region
containing an infinite number of ordered pairs.
Intersecting Regions
B. Solve the system of inequalities by graphing.
y ≤ –x + 1
│x + 1│< 3
The inequality │x + 1│< 3 can be written as x + 1 < 3
and x + 1 > –3, or x < 2 and x > –4.
Intersecting Regions
Graph all of the inequalities on the same coordinate
plane and shade the region or regions that are common
to all.
Answer:
Animation: Use Graphing to Solve the
System of Inequalities
A. Solve the system of inequalities by graphing. What
is the solution to the system of inequalities below?
y ≤ 3x – 3
y>x+1
A.
B.
0%
0%
0%
D
0%
B
D.
A
C.
A
B
C
D
C
1.
2.
3.
4.
B. Solve the system of inequalities by graphing. What
is the solution to the system of inequalities below?
y ≥ –2x – 3
│x + 2│< 1
A.
B.
0%
0%
0%
D
0%
B
D.
A
C.
A
B
C
D
C
1.
2.
3.
4.
Separate Regions
Solve the system of inequalities by graphing.
Graph both inequalities.
The graphs do not overlap, so
the solutions have no points in
common.
Answer: The solution set is Ø.
Solve the system of inequalities by graphing.
0%
0%
0%
0%
D
D.
A
B
C
D
C
1.
2.
3.
4.
B
C.
B.
A
A.
Write and Use a System
of Inequalities
MEDICINE Medical professionals recommend that
patients have a cholesterol level below 200
milligrams per deciliter (mg/dL) of blood and a
triglyceride level below 150 mg/dL. Write and graph a
system of inequalities that represents the range of
cholesterol levels and triglyceride levels for patients.
Let c represent the cholesterol levels in mg/dL. It must be
less than 200 mg/dL. Since cholesterol levels cannot be
negative, we can write this as 0 ≤ c < 200.
Let t represent the triglyceride levels in mg/dL. It must be
less than 150 mg/dL. Since triglyceride levels also
cannot be negative, we can write this as 0 ≤ t < 150.
Write and Use a System
of Inequalities
Graph all of the inequalities. Any ordered pair in the
intersection of the graphs is a solution of the system.
Answer: 0 ≤ c < 200
0 ≤ t < 150
SAFETY The speed limits while driving on the
highway are different for trucks and cars. Cars must
drive between 45 and 65 miles per hour, inclusive.
Trucks are required to drive between 40 and 55 miles
per hour, inclusive. Let c represent the range of
speeds for cars and t represent the range of speeds
for trucks. Which graph represents this situation?
A
B
C
D
0%
0%
0%
D
0%
C
D.
1.
2.
3.
4.
B
C.
B.
A
A.
Find Vertices
Find the coordinates of the vertices of the figure
formed by 2x – y ≥ –1, x + y ≤ 4, and x + 4y ≥ 4.
Graph each inequality. The
intersection of the graphs
forms a triangle.
Answer: The vertices of
the triangle
are at (0, 1),
(4, 0), and (1, 3).
Find the coordinates of the vertices of the figure
formed by x + 2y ≥ 1, x + y ≤ 3, and –2x + y ≤ 3.
A. (–1, 0), (0, 3), and (5, A.
–2) A
B. (–1, 0), (0, 3), and (4, B.
–2) B
C. (–1, 1), (0, 3), and (5, C.
–2) C
0%
0%
D
0%
C
A
0%
B
D. (0, 3), (5, –2), and (1, D.
0) D
Five-Minute Check (over Lesson 3-3)
Main Ideas and Vocabulary
California Standards
Example 1: Bounded Region
Example 2: Unbounded Region
Key Concept: Linear Programming Procedure
Example 3: Linear Programming
• Find the maximum and minimum values of a
function over a region.
• Solve real-world problems using linear programming.
• constraints
• feasible region
• bounded
• vertex
• unbounded
• linear programming
Reinforcement of Standard 2.0 Students solve
systems of linear equations and inequalities (in two or
three variables) by substitution, with graphs, or with
matrices. (Key)
Bounded Region
Graph the following system of inequalities. Name the
coordinates of the vertices of the feasible region.
Find the maximum and minimum values of the
function f(x, y) = 3x – 2y for this region.
x≤5
y≤4
x+y≥2
Step 1 Graph the inequalities.
The polygon formed is a
triangle with vertices at
(–2, 4), (5, –3), and (5, 4).
Bounded Region
Step 2 Use a table to find the maximum and
minimum values of f(x, y). Substitute the
coordinates of the vertices into the function.
← minimum
← maximum
Answer: The vertices of the feasible region are (–2, 4),
(5, –3), and (5, 4). The maximum value is 21 at
(5, –3). The minimum value is –14 at (–2, 4).
Graph the following system of inequalities. What are
the maximum and minimum values of the function
f(x, y) = 4x – 3y for the feasible region of the graph?
x≤4
y≤5
x+y≥6
0%
D
0%
C
0%
B
0%
A
A. maximum: f(4, 5) = 5 A. A
minimum: f(1, 5) = –11
B. maximum: f(4, 2) = 10 B. B
minimum: f(1, 5) = –11
C. maximum: f(4, 2) = 10 C. C
minimum: f(4, 5) = 5
D. maximum: f(1, 5) = –11 D. D
minimum: f(4, 2) = 10
Unbounded Region
Graph the following system of inequalities. Name the
coordinates of the vertices of the feasible region.
Find the maximum and minimum values of the
function f(x, y) = 2x + 3y for this region.
–x + 2y ≤ 2
x – 2y ≤ 4
x + y ≥ –2
Graph the system of
inequalities. There are
only two points of
intersection, (–2, 0)
and (0, –2).
Unbounded Region
The minimum value is –6 at (0, –2). Although f(–2, 0) is
–4, it is not the maximum value since there are other
points that produce greater values. For example, f(2,1) is
7 and f(3, 1) is 10. It appears that because the region is
unbounded, f(x, y) has no maximum value.
Answer: The vertices are at (–2, 0) and (0, –2).
There is no maximum value.
The minimum value is –6 at (0, –2).
Graph the following system of inequalities. What are
the maximum and minimum values of the function
f(x, y) = x + 2y for the feasible region of the graph?
x + 3y ≤ 6
–x – 3y ≤ 9
2y – x ≥ –6
B
C. maximum: f(6, 0) = 6
minimum: no minimum
C.
C
D. maximum: no maximum
minimum: f(0, –3) = –6
D.
D
0%
0%
0%
0%
D
B.
C
B. maximum: f(6, 0) = 6
minimum: f(0, –3) = –6
B
A. A
A
A. maximum: no maximum
minimum: f(6, 0) = 6
Linear Programming Procedure
Step 1
Define the variables.
Step 2
Write a system of inequalities.
Step 3
Graph the system of inequalities.
Step 4
Find the coordinates of the vertices of the
feasible region.
Step 5
Write a function to be maximized or
minimized.
Step 6
Substitute the coordinates of the vertices
into the function.
Step 7
Select the greatest or least result. Answer
the problem.
Linear Programming
LANDSCAPING A landscaping company has crews
who mow lawns and prune shrubbery. The company
schedules 1 hour for mowing jobs and 3 hours for
pruning jobs. Each crew is scheduled for no more
than 2 pruning jobs per day. Each crew’s schedule is
set up for a maximum of 9 hours per day. On the
average, the charge for mowing a lawn is $40 and the
charge for pruning shrubbery is $120. Find a
combination of mowing lawns and pruning shrubs
that will maximize the income the company receives
per day from one of its crews.
Linear Programming
Step 1 Define the variables.
m = the number of mowing jobs
p = the number of pruning jobs
Linear Programming
Step 2 Write a system of inequalities.
Since the number of jobs cannot be negative,
m and p must be nonnegative numbers.
m ≥ 0, p ≥ 0
Mowing jobs take 1 hour. Pruning jobs take
3 hours. There are 9 hours to do the jobs.
1m + 3p ≤ 9
There are no more than 2 pruning jobs a day.
p≤2
Linear Programming
Step 3 Graph the system of inequalities.
Linear Programming
Step 4 Find the coordinates of the vertices of the
feasible region.
From the graph, the vertices are at (0, 2), (3, 2),
(9, 0), and (0, 0).
Step 5 Write the function to be maximized.
The function that describes the income is
f(m, p) = 40m + 120p. We want to find the
maximum value for this function.
Linear Programming
Step 6 Substitute the coordinates of the vertices into
the function.
Step 7 Select the greatest amount.
Linear Programming
Answer: The maximum values are 360 at (3, 2)
and 360 at (9, 0). This means that the
company receives the most money with 3
mowings and 2 prunings or 9 mowings
and 0 prunings.
Animation: Linear Programming
LANDSCAPING A landscaping company has crews
who rake leaves and mulch. The company schedules
2 hours for mulching jobs and 4 hours for raking
jobs. Each crew is scheduled for no more than 2
raking jobs per day. Each crew’s schedule is set up
for a maximum of 8 hours per day. On the average,
the charge for raking a lawn is $50 and the charge for
mulching is $30.
What is a combination of raking leaves and
mulching that will maximize the income the
company receives per day from one of its crews?
B.
B
C. 0 mulching; 4 raking
C.
C
D.
D
A
D. 2 mulching; 0 raking
0%
0%
0%
0%
D
B. 4 mulching; 0 raking
C
A. A
B
A. 0 mulching; 2 raking
Five-Minute Check (over Lesson 3-4)
Main Ideas and Vocabulary
California Standards
Key Concept: System of Equations in
Three Variables
Example 1: One Solution
Example 2: Infinitely Many Solutions
Example 3: No Solution
Example 4: Real-World Example: Write and Solve a
System of Equations
• Solve systems of linear equations in three
variables.
• Solve real-world problems using systems of linear
equations in three variables.
• ordered triple
Standard 2.0 Students solve systems of linear
equations and inequalities (in two or three variables)
by substitution, with graphs, or with matrices. (Key)
System of Equations in Three Variables
One Solution
Solve the system of equations.
5x + 3y + 2z = 2
2x + y – z = 5
x + 4y + 2z = 16
Step 1 Use elimination to make a system of two
equations in two variables.
5x + 3y + 2z = 2
2x + y – z = 5
Multiply by 2.
5x + 3y + 2z = 2 First equation
(+)4x + 2y – 2z = 10 Second equation
9x + 5y
= 12 Add to
eliminate z.
One Solution
5x + 3y + 2z =
2
(–) x + 4y + 2z = 16
4x – y
= –14
First equation
Third equation
Subtract to eliminate z.
Notice that the z terms in each equation have been
eliminated. The result is two equations with the same
two variables x and y.
One Solution
Step 2 Solve the system of two equations.
9x + 5y = 12
4x – y = –14
Multiply by 5
9x + 5y = 12
(+) 20x – 5y = –70
29x
= –58 Add to
eliminate y.
x = –2 Divide by 29.
One Solution
Substitute –2 for x in one of the two equations with two
variables and solve for y.
4x – y = –14
Equation with two variables
4(–2) – y = –14
–8 – y = –14
y=6
Replace x with –2.
Multiply.
Simplify.
The result is x = –2 and y = 6.
One Solution
Step 3 Solve for z using one of the original equations
with three variables.
2x + y – z = 5
Original equation with
three variables
2(–2) + 6 – z = 5
–4 + 6 – z = 5
z = –3
Replace x with –2 and y with 6.
Multiply.
Simplify.
Answer: The solution is (–2, 6, –3). You can
check this solution in the other two
original equations.
What is the solution to the system of equations
shown below?
2x + 3y – 3z = 16
x + y + z = –3
x – 2y – z = –1
A.
A. A
B. (–3, –2, 2)
B.
B
D. (–1, 2, –4)
D.
D
0%
D
C
0%
C
C.
0%
B
C. (1, 2, –6)
A
0%
Infinite Solutions
Solve the system of equations.
2x + y – 3z = 5
x + 2y – 4z = 7
6x + 3y – 9z = 15
Eliminate y in the first and third equations.
Multiply by 3.
2x + y – 3z = 5
6x + 3y – 9z = 15
6x + 3y – 9z = 15
(–)6x + 3y – 9z = 15
0= 0
Infinite Solutions
The equation 0 = 0 is always true. This indicates that
the first and third equations represent the same plane.
Check to see if this plane intersects the second plane.
Multiply by 6.
x + 2y – 4z = 7
6x + 3y – 9z = 15
6x + 12y – 24z = 42
(–)6x + 3y – 9z = 15
9y – 15z = 27
Divide by the GCF, 3.
3y – 5z = 9
Answer: The planes intersect in a line. So, there is an
infinite number of solutions.
What is the solution to the system of equations
shown below?
x + y – 2z = 3
–3x – 3y + 6z = –9
2x + y – z = 6
A. (1, 2, 0)
A. A
B. (2, 2, 0)
B.
B
D
D. no solution
0%
D
D.
0%
C
C
0%
B
C. infinite number of C.
solutions
A
0%
No Solution
Solve the system of equations.
3x – y – 2z = 4
6x + 4y + 8z = 11
9x + 6y + 12z = –3
Eliminate x in the last two equations.
Multiply by 3.
6x + 4y + 8z = 11
18x + 12y + 24z = 33
9x + 6y + 12z = –3 (–) 18x + 12y + 24z = –6
Multiply by 2.
0 = 39
Answer: The equation 0 = 39 is never true. So,
there is no solution of this system.
What is the solution to the system of equations
shown below?
3x + y – z = 5
–15x – 5y + 5z = 11
x+y+z=2
A. (0, 6, 1)
A. A
B. (1, 0, –2)
B.
B
D. no solution
D.
D
0%
D
0%
C
0%
B
C. infinite number ofC.solutions
C
A
0%
Write and Solve a
System of Equations
SPORTS There are 49,000 seats in a sports stadium.
Tickets for the seats in the upper level sell for $25,
the ones in the middle level cost $30, and the ones in
the bottom level are $35 each. The number of seats
in the middle and bottom levels together equals the
number of seats in the upper level. When all of the
seats are sold for an event, the total revenue is
$1,419,500. How many seats are there in each level?
Explore Read the problem and define the variables.
u = number of seats in the upper level
m = number of seats in the middle level
b = number of seats in the bottom level
Write and Solve a
System of Equations
Plan
There are 49,000 seats.
u + m + b = 49,000
When all the seats are sold, the revenue is
1,419,500. Seats cost $25, $30, and $35.
25u + 30m + 35b = 1,419,500
The number of seats in the middle and bottom
levels together equal the number of seats in the
upper level.
m+b=u
Write and Solve a
System of Equations
Solve Substitute u = m + b in each of the first
two equations.
(m + b) + m + b = 49,000
2m + 2b = 49,000
m + b = 24,500
Replace u with m + b.
Simplify.
Divide by 2.
25(m + b) + 30m + 35b = 1,419,500 Replace u with
m + b.
25m + 25b + 30m + 35b = 1,419,500 Distributive
Property
55m + 60b = 1,419,500 Simplify.
Write and Solve a
System of Equations
Now, solve the system of two equations in two variables.
Multiply by 55.
m + b = 24,500
55m + 60b = 1,419,500
55m + 55b = 1,347,500
(–) 55m + 60b = 1,419,500
–5b =
–72,000
b = 14,400
Write and Solve a
System of Equations
Substitute 14,400 for b in one of the equations with two
variables and solve for m.
m + b = 24,500
m + 14,400 = 24,500
m = 10,100
Equation with two variables
b = 14,400
Subtract 14,400 from
each side.
Write and Solve a
System of Equations
Substitute 14,400 for b and 10,100 for m in one of the
original equations with three variables.
m+b =u
10,100 + 14,400 = u
24,500 = u
Equation with three
variables
m = 10,100, b = 14,400
Add.
Answer: There are 24,500 upper level, 10,100 middle
level, and 14,400 bottom level seats.
Write and Solve a
System of Equations
Check Check to see if all the criteria are met.
24,500 + 10,100 + 14,400 = 49,000
The number of seats in the middle and bottom
levels equals the number of seats in the
upper level.
10,100 + 14,400 = 24,500
When all of the seats are sold, the revenue
is $1,419,500.
24,500($25) + 10,100($30) +
14,400($35) = $1,419,500
BUSINESS The school store sells pens, pencils, and
paper. The pens are $1.25 each, the pencils are $0.50
each, and the paper is $2 per pack. Yesterday the
store sold 25 items and earned $32. The number of
pens sold equaled the number of pencils sold plus
the number of packs of paper sold minus 5. How
many of each item did the store sell?
Interactive Lab: Solving Systems of Linear
Equations and Inequalities
How many of each item did the store sell?
A. pens: 5; pencils: 10;A. A
paper: 10
C
D
A
D. pens: 11; pencils: 2; D.
paper: 12
0%
0%
0%
0%
D
C. pens: 10; pencils: 7;C.
paper: 8
C
B
B
B. pens: 8; pencils: 7; B.
paper: 10
Five-Minute Checks
Image Bank
Math Tools
Use Graphing to Solve the System of
Inequalities
Linear Programming
Solving Systems of Linear Equations
and Inequalities
Lesson 3-1 (over Chapter 2)
Lesson 3-2 (over Lesson 3-1)
Lesson 3-3 (over Lesson 3-2)
Lesson 3-4 (over Lesson 3-3)
Lesson 3-5 (over Lesson 3-4)
To use the images that are on the
following three slides in your own
presentation:
1. Exit this presentation.
2. Open a chapter presentation using a
full installation of Microsoft® PowerPoint®
in editing mode and scroll to the Image
Bank slides.
3. Select an image, copy it, and paste it
into your presentation.
(over Chapter 2)
Find the domain (D) and range (R) of the relation
{(–4, 1), (0, 0), (1, –4), (2, 0), (–2, 0)}. Determine
whether the relation is a function.
A. D = {–4, –2, 0, 1, 2}, A. A
R = {–4, 0, 1}; no
B. D = {–4, 0, 1},
B. B
R = {–4, –2, 0, 1, 2}; no
0%
0%
D
0%
C
D. D = {–4, 0, 1},
D. D
R = {–4, –2, 0, 1, 2}; yes
0%
B
C
A
C. D = {–4, –2, 0, 1, 2}, C.
R = {–4, 0, 1}; yes
(over Chapter 2)
Find the value of f(4) for f(x) = 8 – x – x2.
B.
B
C. –4
C.
C
D. –12
D.
D
A
0%
0%
0%
0%
D
B. 12
C
A. A
B
A. 20
(over Chapter 2)
Find the slope of the line that passes through
(5, 7) and (–1, 0).
A. A
A.
C
D.
D.
D
A
0%
0%
0%
0%
D
C.
C
C.
B
B
B.
B.
(over Chapter 2)
Write an equation in slope-intercept form for the line
that has x-intercept –3 and y-intercept 6.
B.
B
C. y = –3x + 6
C.
C
D. y = x + 3
D.
D
A
0%
0%
0%
0%
D
B. y = 3x – 6
C
A. A
B
A. y = 2x + 6
(over Chapter 2)
The Math Club is using the prediction equation
y = 1.25x + 10 to estimate the number of members it
will have, where x represents the number of years
the club has been in existence. About how many
members does the club expect to have in its fifth
year?
A. 12
A. A
B. 15
B.
D. 19
D.
D
0%
D
C.0% C
0%
C
A
0%
B
C. 16
B
(over Chapter 2)
Which function has the greatest value for f(–2)?
A. f(x) = x
B
D.
f(x) = – |x|
C.
C
D.
D
A
0%
0%
0%
0%
D
B.
C
C. f(x) = |x| –1
A. A
B
B. f(x) = [x]
(over Lesson 3-1)
Which choice shows a graph of the solution of
this system?
y = 3x – 2
y = –3x + 2
0%
0% 1.0% A 0%
2.
3.
4.
B
C
D
D
D.
C
C.
B
B.
A
A.
(over Lesson 3-1)
State whether the graph of the system of equations
is consistent and independent, consistent and
dependent, or inconsistent?
2x + y = 6
3y = –6x + 6
A. A
A. consistent and
independent
B. B
B. consistent and
dependent
C. C
C. inconsistent
0%
D
0%
C
A
D. none of the above
D. D
0%
B
0%
(over Lesson 3-1)
Which equation is inconsistent with 4x + 5y = 30?
A. A
C.
C
D. 5y = 4x – 30
D.
D
0%
0%
0%
0%
D
C.
C
B
A
B. 5y = –4x – 30 B.
B
A.
(over Lesson 3-2)
Solve the system of equations by using substitution.
x – 4y = –12
3x + 2y = 20
B
C. no solution
C.
C
0%
D. infinitely many solutions
D. D
0%
0%
0%
D
B.
C
B. (8, 5)
B
A. A
A
A. (4, 4)
(over Lesson 3-2)
Solve the system of equations by using substitution.
x + 2y = 6
2x + 4y = 15
B
C. no solution
C.
C
0%
D. infinitely many solutions
D. D
0%
0%
0%
D
B.
C
B. (3.5, 2)
B
A. A
A
A. (3, 1.5)
(over Lesson 3-2)
Solve the system of equations by using elimination.
2x + 5y = 9
–2x + 8y = 4
B
C. no solution
C.
C
0%
D. infinitely many solutions
D. D
0%
0%
0%
D
B.
C
B. (2, 1)
B
A. A
A
A. (–2, 0)
(over Lesson 3-2)
Solve the system of equations by using elimination.
x + 2y = 7
14 – 4y = 2x
B
C. no solution C.
C
0%
D. infinitely many
D. solutions
D
0%
0%
0%
D
B.
C
B. (–5, –6)
B
A. A
A
A. (0, 0)
(over Lesson 3-2)
Two times one number minus a second number is 7.
The second number is three more than the first
number. What are the two numbers?
A. A
A. 7 and 10
C.
C
D.
D
A
D. 7 and 20
0%
0%
0%
0%
D
B
C
C. 13 and 16
B.
B
B. 10 and 13
(over Lesson 3-3)
Which choice shows a graph of the solution of
the system of inequalities?
x ≤ –2
y>3
A.
B.
0%
0%
D
0%
C
0%
A
B
C
D
B
D.
A
C.
1.
2.
3.
4.
(over Lesson 3-3)
Which choice shows a graph of the solution of
the system of inequalities?
y ≤ 3x + 2
y > –x
A.
B.
0%
0%
D
0%
A
B
C
D
C
0%
B
D.
A
C.
1.
2.
3.
4.
(over Lesson 3-3)
Find the coordinates of the vertices of the figure
formed by the system of inequalities.
x≥0
y≤0
–3x + y = –6
A. (0, 0), (2, 0), (–6, 0) A. A
B
C
D. (0, 0), (0, –2), (0, –6) D.
D
A
C. (0, 0), (2, 0), (0, –6) C.
0%
0%
0%
D
0%
C
B.
B
B. (0, 0), (0, 2), (0, 6)
(over Lesson 3-3)
Find the coordinates of the vertices of the figure
formed by the system of inequalities.
y≤3
y≥0
x≥0
2y + 3x ≤ 12
A. A
A. (0, 0), (0, 6), (2, 3), (4, 0)
B. B
D. (0, 0), (0, 3), (2, 3), (4, 0)
0%
0%
D
0%
C
A
C. (0, 0), (3, 0), (0, 6), (2, 3)
D. D
0%
B
B. (0, 0), (3, 0), (3, 2), (4, 0)
C. C
(over Lesson 3-3)
Which point is not the solution of the system?
|x|  2
y3
A. A
A. (0, 4)
B.
B
C.
C
B. (–3, 5)
0%
D
D
0%
C
D.
0%
B
D. (1, 4)
0%
A
C. (–2, 5)
(over Lesson 3-4)
Which choice shows a graph of the system of
inequalities and the coordinates of the vertices of the
feasible region?
1≤x≤4
A.
B.
y≥x
y ≤ 2x + 3
0%
0%
D
0%
C
0%
B
D.
A
B
C
D
A
C.
1.
2.
3.
4.
(over Lesson 3-4)
The vertices of a feasible region are (0, 0), (0, 6),
(4, 2), and (5, 5). What are the maximum and
minimum values of the function f(x, y) = 3x – 5y
over this region?
A. minimum: f(0, 0) = 0A. A
0%
D
0%
C
0%
B
0%
A
maximum: f(5, 5) = –10
B. minimum: f(0, 0) = 0B. B
maximum: f(4, 2) = 2
C. C
C. minimum: f(0, 6) = –30
maximum: f(4, 2) = 2
D. D
D. minimum: f(0, 6) = –30
maximum: f(5, 5) = –10
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