6-4 Elimination Using Multiplication
Use elimination to solve each system of
equations.
1. 2x − y = 4
7x + 3y = 27
SOLUTION: Notice that if you multiply the first equation by 3, the
coefficients of the y–terms are additive inverses.
Now, substitute 3 for x in either equation to find y.
The solution is (–3, 1).
3. 4x + 2y = −14
5x + 3y = −17
SOLUTION: Notice that if you multiply the first equation by 3 and
the second equation by –2, the coefficients of the y–
terms are additive inverses.
Now, substitute –4 for x in either equation to find y.
The solution is (–4, 1).
The solution is (3, 2).
2. 2x + 7y = 1
x + 5y = 2
SOLUTION: Notice that if you multiply the second equation by –2,
the coefficients of the x–terms are additive inverses.
Now, substitute 1 for y in either equation to find x.
4. 9a − 2b = −8
−7a + 3b = 12
SOLUTION: Notice that if you multiply the first equation by 3 and
the second equation by 2, the coefficients of the y–
terms are additive inverses.
Now, substitute 0 for a in either equation to find b.
The solution is (0, 4).
The solution is (–3, 1).
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3. 4x + 2y = −14
5x + 3y = −17
5. KAYAKING A kayaking group with a guide
travels 16 miles downstream, stops for a meal, and
Page 1
then travels 16 miles upstream. The speed of the
current remains constant throughout the trip. Find the
speed of the kayak in still water.
6-4 Elimination
Using Multiplication
The solution is (0, 4).
So, the speed of the kayak is 6 mph.
5. KAYAKING A kayaking group with a guide
travels 16 miles downstream, stops for a meal, and
then travels 16 miles upstream. The speed of the
current remains constant throughout the trip. Find the
speed of the kayak in still water.
SOLUTION: Let x represent the speed of the kayak and y
represent the speed of river.
The rate to travel down river would be with the
current so the speed of kayak and speed of river
are added. The rate to travel against the current
would be the difference between the speed of the
kayak and speed of the river. with current
against current
speed of speed of time
d
kayak
river (hours)
d = rt
x
y
2
16 16 = 2(x + y)
x
y
4
16 16 = 4(x + y)
6. PODCASTS Steve subscribed to 10 podcasts for a
total of 340 minutes. He used his two favorite tags,
Hobbies and Recreation and Soliloquies. Each of the
Hobbies and Recreation episodes lasted about 32
minutes. Each Soliloquies episodes lasted 42 minutes.
To how many of each tag did Steve subscribe?
SOLUTION: Let x represent the number of Hobbies and
Recreation podcasts and y represent the number of
Soliloquies podcasts.
Since Steve subscribed to 10 podcasts, then the
number of Hobbies and Recreation and number of
Soliloquies would equal 10. Each Hobbies and
Recreation podcast talks 32 minutes. SO the total
minutes for Hobbies and Recreation podcast is 32x.
Each Soliloquies podcast talks 42 minutes. So the
total minutes for Soliloquies podcast is 42x. Add
these and equal to the total minutes of 340. Notice that if you multiply the first equation by –32,
the coefficients of the x–terms are additive inverses.
Now, substitute 2 for y in either equation to find x.
Notice that the coefficients of the y–terms are
additive inverses, so add the equations.
Steve subscribed to 2 Soliloquies and 8 Hobby and
Recreation podcasts.
So, the speed of the kayak is 6 mph.
6. PODCASTS Steve subscribed to 10 podcasts for a
total of 340 minutes. He used his two favorite tags,
Hobbies and Recreation and Soliloquies. Each of the
Hobbies and Recreation episodes lasted about 32
minutes. Each Soliloquies episodes lasted 42 minutes.
To how many of each tag did Steve subscribe?
Use elimination to solve each system of
equations.
7. x + y = 2
−3x + 4y = 15
SOLUTION: Notice that if you multiply the first equation by 3, the
coefficients of the x–terms are additive inverses.
SOLUTION: eSolutions
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by Cognero
Let xManual
represent
the number
of
Hobbies and
Recreation podcasts and y represent the number of
Soliloquies podcasts.
Page 2
6-4 Elimination
Using
Steve subscribed
to 2Multiplication
Soliloquies and 8 Hobby and
Recreation podcasts.
Use elimination to solve each system of
equations.
7. x + y = 2
−3x + 4y = 15
SOLUTION: Notice that if you multiply the first equation by 3, the
coefficients of the x–terms are additive inverses.
Now, substitute 3 for y in either equation to find x.
The solution is (–2, 6).
9. x + 5y = 17
−4x + 3y = 24
SOLUTION: Notice that if you multiply the first equation by 4, the
coefficients of the x–terms are additive inverses.
Now, substitute 4 for y in either equation to find x.
The solution is (–3, 4).
The solution is (–1, 3).
10. 6x + y = −39
3x + 2y = −15
8. x − y = −8
7x + 5y = 16
SOLUTION: Notice that if you multiply the first equation by 5, the
coefficients of the y–terms are additive inverses.
SOLUTION: Notice that if you multiply the first equation by –2,
the coefficients of the y–terms are additive inverses.
Now, substitute –2 for x in either equation to find y.
Now, substitute –7 for x in either equation to find y.
The solution is (–7, 3).
The solution is (–2, 6).
9. x + 5y = 17
−4x + 3y = 24
SOLUTION: Notice that if you multiply the first equation by 4, the
coefficients of the x–terms are additive inverses.
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11. 2x + 5y = 11
4x + 3y = 1
SOLUTION: Notice that if you multiply the first equation by –2,
the coefficients of the x–terms are additive inverses.
Page 3
6-4 Elimination
Using Multiplication
The solution is (–7, 3).
11. 2x + 5y = 11
4x + 3y = 1
The solution is (0, 2).
13. 3x + 4y = 29
6x + 5y = 43
SOLUTION: Notice that if you multiply the first equation by –2,
the coefficients of the x–terms are additive inverses.
SOLUTION: Notice that if you multiply the first equation by –2,
the coefficients of the x–terms are additive inverses.
Now, substitute 3 for y in either equation to find x.
Now, substitute 5 for y in either equation to find x.
The solution is (–2, 3).
The solution is (3, 5).
12. 3x − 3y = −6
−5x + 6y = 12
SOLUTION: Notice that if you multiply the first equation by 2, the
coefficients of the y–terms are additive inverses.
Now, substitute 0 for x in either equation to find y.
14. 8x + 3y = 4
−7x + 5y = −34
SOLUTION: Notice that if you multiply the first equation by 7 and
the second equation by 8, the coefficients of the x–
terms are additive inverses.
Now, substitute –4 for y in either equation to find x.
The solution is (0, 2).
13. 3x + 4y = 29
6x + 5y = 43
SOLUTION: Notice that if you multiply the first equation by –2,
the coefficients of the x–terms are additive inverses.
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The solution is (2, –4).
15. 8x + 3y = −7
7x + 2y = −3
Page 4
SOLUTION: Notice that if you multiply the first equation by –2
6-4 Elimination
Using Multiplication
The solution is (2, –4).
15. 8x + 3y = −7
7x + 2y = −3
The solution is (1, –5).
16. 4x + 7y = −80
3x + 5y = −58
SOLUTION: Notice that if you multiply the first equation by –2
and the second equation by 3, the coefficients of the
y–terms are additive inverses.
SOLUTION: Notice that if you multiply the first equation by –3
and the second equation by 4, the coefficients of the
x–terms are additive inverses.
Now, substitute 1 for x in either equation to find y.
Now, substitute –8 for y in either equation to find x.
The solution is (1, –5).
16. 4x + 7y = −80
3x + 5y = −58
SOLUTION: Notice that if you multiply the first equation by –3
and the second equation by 4, the coefficients of the
x–terms are additive inverses.
Now, substitute –8 for y in either equation to find x.
The solution is (–6, –8).
17. 12x − 3y = −3
6x + y = 1
SOLUTION: Notice that if you multiply the second equation by –2,
the coefficients of the x–terms are additive inverses.
Now, substitute 1 for y in either equation to find x.
The solution is (0, 1).
The solution is (–6, –8).
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17. 12x − 3y = −3
6x + y = 1
18. −4x + 2y = 0
10x + 3y = 8
SOLUTION: Page 5
Notice that if you multiply the first equation by –3
and multiply the second equation by 2, the
6-4 Elimination
Using Multiplication
The solution is (0, 1).
18. −4x + 2y = 0
10x + 3y = 8
SOLUTION: Notice that if you multiply the first equation by –3
and multiply the second equation by 2, the
coefficients of the y–terms are additive inverses.
Now, substitute
for x in either equation to find y.
The solution is
.
19. NUMBER THEORY Seven times a number plus
three times another number equals negative one. The
sum of the two numbers is negative three. What are
the numbers?
SOLUTION: Let x represent one number and y represent the
second number.
Notice that if you multiply the second equation by –3,
the coefficients of the y–terms are additive inverses.
Now, substitute 2 for x in either equation to find y.
The solution is
.
19. NUMBER THEORY Seven times a number plus
three times another number equals negative one. The
sum of the two numbers is negative three. What are
the numbers?
SOLUTION: Let x represent one number and y represent the
second number.
Notice that if you multiply the second equation by –3,
the coefficients of the y–terms are additive inverses.
Now, substitute 2 for x in either equation to find y.
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So, the two numbers are 2 and –5.
20. FOOTBALL A field goal is 3 points and the extra
point after a touchdown is 1 point. In a recent post–
season, Adam Vinatieri of the Indianapolis Colts
made a total of 21 field goals and extra point kicks
for 49 points. Find the number of field goals and
extra points that he made.
SOLUTION: Let x represent the number of field goals and y
represent the number of extra point kicks.
Notice that if you multiply the first equation by –1,
the coefficients of the y–terms are additive inverses.
Now, substitute 14 for x in either equation to findPage
y. 6
So, he made 14 field goals and 7 extra point kicks.
6-4 Elimination
Using Multiplication
So, the two numbers are 2 and –5.
20. FOOTBALL A field goal is 3 points and the extra
point after a touchdown is 1 point. In a recent post–
season, Adam Vinatieri of the Indianapolis Colts
made a total of 21 field goals and extra point kicks
for 49 points. Find the number of field goals and
extra points that he made.
Use elimination to solve each system of
equations.
21. 2.2x + 3y = 15.25
4.6x + 2.1y = 18.325
SOLUTION: Notice that if you multiply the first equation by –2.1
and the second equation by 3, the coefficients of the
y–terms are additive inverses.
SOLUTION: Let x represent the number of field goals and y
represent the number of extra point kicks.
Notice that if you multiply the first equation by –1,
the coefficients of the y–terms are additive inverses.
Now, substitute 2.5 for x in either equation to find y.
Now, substitute 14 for x in either equation to find y.
So, the solution is (2.5, 3.25).
So, he made 14 field goals and 7 extra point kicks.
22. −0.4x + 0.25y = −2.175
2x + y = 7.5
SOLUTION: Notice that if you multiply the second equation by –
0.25, the coefficients of the y–terms are additive
inverses.
Use elimination to solve each system of
equations.
21. 2.2x + 3y = 15.25
4.6x + 2.1y = 18.325
SOLUTION: Notice that if you multiply the first equation by –2.1
and the second equation by 3, the coefficients of the
y–terms are additive inverses.
Now, substitute 4.5 for x in either equation to find y.
Now, substitute 2.5 for x in either equation to find y.
So, the solution is (4.5, –1.5).
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Page 7
23. 6-4 Elimination
Using Multiplication
So, the solution is (4.5, –1.5).
So, the solution is
.
24. 23. SOLUTION: Notice that if you multiply the second equation by –8,
the coefficients of the y–terms are additive inverses.
SOLUTION: Notice that if you multiply the second equation by –
12, the coefficients of the y–terms are additive
inverses.
Now, substitute 3 for x in either equation to find y.
Now, substitute
for x in either equation to find y.
So, the solution is
.
So, the solution is
24. SOLUTION: Notice that if you multiply the second equation by –
12, the coefficients of the y–terms are additive
inverses.
.
25. CCSS MODELING A staffing agency for in-home
nurses and support staff places necessary personal at
locations on a daily basis. Each nurse placed works
240 minutes per day at a daily rate of $90. Each
support staff employee works 360 minutes per day at
a daily rate of $120.
a. On a given day, 3000 total minutes are worked by
the nurses and support staff that were placed. Write
an equation that represents this relationship. b. On the same day, $1050 of total wages were
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eSolutions
Now, substitute
for x in either equation to find y.
earned by the placed nurses and support staff. Write
an equation that represents this relationship. Page 8
c. Solve the system of equations, and interpret the
26. 6-4 Elimination Using Multiplication
So, the solution is
.
The graphs of x + 2y = 6 and 2x + y
= 9 contain two of the sides of a triangle. A vertex of
the triangle is at the intersection of the graphs.
a. What are the coordinates of the vertex?
25. CCSS MODELING A staffing agency for in-home
nurses and support staff places necessary personal at
locations on a daily basis. Each nurse placed works
240 minutes per day at a daily rate of $90. Each
support staff employee works 360 minutes per day at
a daily rate of $120.
a. On a given day, 3000 total minutes are worked by
the nurses and support staff that were placed. Write
an equation that represents this relationship. b. On the same day, $1050 of total wages were
earned by the placed nurses and support staff. Write
an equation that represents this relationship. b. Draw the graph of the two lines. Identify the
vertex of the triangle.
c. The line that forms the third side of the triangle is
the line x − y = −3. Draw this line on the previous
graph.
d. Name the other two vertices of the triangle.
SOLUTION: a. Notice that if you multiply the first equation by –2,
the coefficients of the x–terms are additive inverses.
c. Solve the system of equations, and interpret the
solution in context of the situation.
SOLUTION: a. Let n represent the number of nurse’s minutes
and s represent the number of support staff minutes.
240n + 360s = 3000
Now, substitute 1 for y in either equation to find x.
b. 90n + 120s = 1050
c. Notice that if you multiply the second equation by
–3, the coefficients of the s–terms are additive
inverses.
The vertex is (4, 1).
b. Rewrite each equation in slope-intercept form and
then graph on the same coordinate plane.
Now, substitute 5 for n in either equation to find s.
(5, 5). 26. GEOMETRY The graphs of x + 2y = 6 and 2x + y
= 9 contain two of the sides of a triangle. A vertex of
the triangle is at the intersection of the graphs.
a. What are the coordinates of the vertex?
b. Draw
graph by
of Cognero
the two
eSolutions
Manualthe
- Powered
vertex of the triangle.
lines. Identify the
The vertex is (4, 1).
c. Rewrite the equation in slope-intercept form and
then graph the line on the same coordinate plane with
the first two equations.
Page 9
The vertex is (4, 1).
c. Rewrite theUsing
equation
in slope-intercept form and
6-4 Elimination
Multiplication
then graph the line on the same coordinate plane with
the first two equations.
So, the vertex is (2, 5).
27. ENTERTAINMENT At an entertainment center,
two groups of people bought batting tokens and
miniature golf games, as shown in the table.
a. Define the variables, and write a system of linear
equations from this situation.
d. To find the other two vertices, use elimination to
find the solutions to the other two systems of
equations created by the three lines.
b. Solve the system of equations, and explain what
the solution represents.
SOLUTION: a. Let x = the cost of a batting token and let y = the
cost of a miniature golf game; 16x + 3y = 30 and 22x
+ 5y = 43.
Substitute 3 for y in either equation to find the value
of x.
b. Notice that if you multiply the first equation by –5
and the multiply the second equation by 3, the
coefficients of the y–terms are additive inverses.
Now, substitute 1.5 for x in either equation to find y.
So, the vertex is (0, 3).
Substitute 2 for x in either equation to find the value
of y .
The solution is (1.5, 2). A batting token costs $1.50
and a game of miniature golf costs $2.00.
So, the vertex is (2, 5).
27. ENTERTAINMENT At an entertainment center,
two groups of people bought batting tokens and
miniature golf games, as shown in the table.
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28. TESTS Mrs. Henderson discovered that she had
accidentally reversed the digits of a test score and
did not give a student 36 points. Mrs. Henderson told
the student that the sum of the digits was 14 and
agreed to give the student his correct score plus
extra credit if he could determine his actual score.
What was his correct score?
SOLUTION: Page 10
Let x represent the tens digit and y represent the
ones digit.
6-4 Elimination
The solution Using
is (1.5,Multiplication
2). A batting token costs $1.50
and a game of miniature golf costs $2.00.
28. TESTS Mrs. Henderson discovered that she had
accidentally reversed the digits of a test score and
did not give a student 36 points. Mrs. Henderson told
the student that the sum of the digits was 14 and
agreed to give the student his correct score plus
extra credit if he could determine his actual score.
What was his correct score?
SOLUTION: Let x represent the tens digit and y represent the
ones digit.
The sum of the digits is 14, so x + y = 14. Since
Mrs. Henderson switched the digits, she gave the
students 10y + x points, when they actually earned
10x + y points.The difference in points owed and
points given is 36. So, the correct score is 95.
29. REASONING Explain how you could recognize a
system of linear equations with infinitely many
solutions.
SOLUTION: The system of equations 2x - 5y = 14 and 12x - 30y
= 84 will have infinitely many solutions. You can
solve by substitution or elimination and get a true
statement, such as, 0 = 0. You can also notice that
12x - 30y = 84 is 6(2x - 5y = 14). The system of
equations will have infinitely many solutions
whenever one of the equations is a multiple of the
other.
30. ERROR ANALYSIS Jason and Daniela are
solving a system of equations. Is either of them
correct? Explain your reasoning.
Notice that if you multiply the first equation by 9 the
x–terms are the same, so subtract the equations.
Now, substitute 5 for y in either equation to find x.
So, the correct score is 95.
29. REASONING Explain how you could recognize a
system of linear equations with infinitely many
solutions.
SOLUTION: The system of equations 2x - 5y = 14 and 12x - 30y
= 84 will have infinitely many solutions. You can
solve by substitution or elimination and get a true
statement, such as, 0 = 0. You can also notice that
12x - 30y = 84 is 6(2x - 5y = 14). The system of
equations will have infinitely many solutions
whenever one of the equations is a multiple of the
other.
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30. ERROR ANALYSIS Jason and Daniela are
solving a system of equations. Is either of them
SOLUTION: Jason is correct. In order to eliminate the r–terms,
you must multiply the second equation by 2 and then
subtract, or multiply the equation by −2 and then add.
When Daniela subtracted the equations, she should
have gotten r + 16t = 18, instead of r = 18. The tterm should not be eliminated. She needs to find
multiples of the equations that have the same
coefficient or opposite coefficients for either r or t
before adding or subtracting the equations.
31. OPEN ENDED Write a system of equations that
can be solved by multiplying one equation by −3 and
then adding the two equations together.
SOLUTION: Sample answer: 2x + 3y = 6, 4x + 9y = 5
Page 11
have gotten r + 16t = 18, instead of r = 18. The tterm should not be eliminated. She needs to find
multiples of the equations that have the same
or t
6-4 Elimination
Using Multiplication
coefficient or opposite coefficients for either r
before adding or subtracting the equations.
31. OPEN ENDED Write a system of equations that
can be solved by multiplying one equation by −3 and
then adding the two equations together.
SOLUTION: Sample answer: 2x + 3y = 6, 4x + 9y = 5
Substitute
y.
32. CHALLENGE The solution of the system 4x + 5y
= 2 and 6x − 2y = b is (3, a). Find the values of a
and b. Discuss the steps that you used.
SOLUTION: We already know that the value of x is 3. We can
use that to find y.
Substitute 3 for x and solve for y.
for x in either equation and solve for
32. CHALLENGE The solution of the system 4x + 5y
= 2 and 6x − 2y = b is (3, a). Find the values of a
and b. Discuss the steps that you used.
SOLUTION: We already know that the value of x is 3. We can
use that to find y.
Substitute 3 for x and solve for y.
So y = –2, which means that a = –2. Now substitute
–2 for y and 3 for x into the second equations and
solve for b.
a = −2, b = 22
33. WRITING IN MATH Why is substitution
sometimes more helpful than elimination, and vice
versa?
So y = –2, which means that a = –2. Now substitute
–2 for y and 3 for x into the second equations and
solve for b.
a = −2, b = 22
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33. WRITING IN MATH Why is substitution
SOLUTION: Sample answer: It is more helpful to use substitution
when one of the variables has a coefficient of 1 or if
a coefficient can be reduced to 1 without turning
other coefficients into fractions. Otherwise,
elimination is more helpful because
it will avoid the use of fractions when solving the
system.
34. What is the solution of this system of equations?
A (3, 3)
B (−3, 3)
C (−3, 1)
D (1, −3)
Page 12
SOLUTION: Notice that if you multiply the second equation by 2,
a coefficient can be reduced to 1 without turning
other coefficients into fractions. Otherwise,
elimination is more helpful because
it will avoid the
use Multiplication
of fractions when solving the
6-4 Elimination
Using
system.
34. What is the solution of this system of equations?
A (3, 3)
B (−3, 3)
C (−3, 1)
D (1, −3)
So, the solution is (–3, 1) and the correct choice is C.
35. A buffet has one price for adults and another for
children. The Taylor family has two adults and three
children, and their bill was $40.50. The Wong family
has three adults and one child. Their bill was $38.
Which system of equations could be used to
determine the price for an adult and for a child?
SOLUTION: Notice that if you multiply the second equation by 2,
the coefficients of the x–terms are additive inverses.
Now, substitute 1 for y in either equation to find x.
So, the solution is (–3, 1) and the correct choice is C.
35. A buffet has one price for adults and another for
children. The Taylor family has two adults and three
children, and their bill was $40.50. The Wong family
has three adults and one child. Their bill was $38.
Which system of equations could be used to
determine the price for an adult and for a child?
SOLUTION: If x is the adult ticket price and y is the child’s ticket
price, then the Taylor family’s price would be 2
adults (2x) and 3 children (3y). That means that their
price could be represented by
. Thus,
you can eliminate choices F and J. The Wong family
has 3 adults (3x) and 1 child (1y), so their price could
be represented by
. Therefore, the
correct choice is G.
36. SHORT RESPONSE A customer at a paint store
has ordered 3 gallons of ivy green paint. Melissa
mixes the paint in a ratio of 3 parts blue to one part
yellow. How many quarts of blue paint does she use?
SOLUTION: Let b represent the number of quarts of blue paint
mixed and y represent the number of quarts of
yellow paint mixed . If the customer ordered 3
gallons of paint, Melissa needs to mix 3(4), or 12 quarts. Write a system of equations that describes
the problem.
first equation: b + y = 12
second equation: If the first equation is solved for y , you get y = 12 b. Substitute this into the second equation and solve
for b.
SOLUTION: If x is the adult ticket price and y is the child’s ticket
price, then the Taylor family’s price would be 2
adults (2x) and 3 children (3y). That means that their
price could be represented by
. Thus,
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you can eliminate choices F and J. The Wong family
has 3 adults (3x) and 1 child (1y), so their price could
be represented by
. Therefore, the
Page 13
price could be represented by
. Thus,
you can eliminate choices F and J. The Wong family
has 3 adults (3x) and 1 child (1y), so their price could
be represented
by Multiplication
. Therefore, the
6-4 Elimination
Using
correct choice is G.
36. SHORT RESPONSE A customer at a paint store
has ordered 3 gallons of ivy green paint. Melissa
mixes the paint in a ratio of 3 parts blue to one part
yellow. How many quarts of blue paint does she use?
SOLUTION: Let b represent the number of quarts of blue paint
mixed and y represent the number of quarts of
yellow paint mixed . If the customer ordered 3
gallons of paint, Melissa needs to mix 3(4), or 12 quarts. Write a system of equations that describes
the problem.
first equation: b + y = 12
second equation: If the first equation is solved for y , you get y = 12 b. Substitute this into the second equation and solve
for b.
Therefore, she will need 9 quarts of blue paint.
37. PROBABILITY The table shows the results of a
number cube being rolled. What is the experimental
probability of rolling a 3?
Outcome
1
2
3
4
5
6
Frequency
4
8
2
0
5
1
A B C 0.2
D 0.1
SOLUTION: Find the total number of outcomes = 4 + 8 + 2 + 5 +
1 = 20.
The experimental probability is
Therefore, she will need 9 quarts of blue paint.
37. PROBABILITY The table shows the results of a
number cube being rolled. What is the experimental
probability of rolling a 3?
Outcome
1
2
3
4
5
6
Frequency
4
8
2
0
5
1
.
So, the correct choice is D.
Use elimination to solve each system of
equations.
38. f + g = −3
f −g=1
SOLUTION: Notice the coefficients for the g–terms are the
opposite, so add the equations.
A B Now, substitute –1 for f in either equation to find g.
C 0.2
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D 0.1
Page 14
1 = 20.
The experimental probability is
.
6-4 Elimination
Using Multiplication
So, the correct choice is D.
Use elimination to solve each system of
equations.
38. f + g = −3
f −g=1
SOLUTION: Notice the coefficients for the g–terms are the
opposite, so add the equations.
The solution is (–1, –2).
39. 6g + h = −7
6g + 3h = −9
SOLUTION: Notice the coefficients for the g–terms are the same,
multiply equation 2 by –1, then add the equations to
find h.
Now, substitute –1 for f in either equation to find g.
Now, substitute –1 for h in either equation to find g.
The solution is (–1, –2).
39. 6g + h = −7
6g + 3h = −9
SOLUTION: Notice the coefficients for the g–terms are the same,
multiply equation 2 by –1, then add the equations to
find h.
The solution is (–1, –1).
40. 5j + 3k = −9
3j + 3k = −3
SOLUTION: Notice the coefficients for the k–terms are the same,
multiply equation 2 by –1 and add the equations to
find j .
Now, substitute –1 for h in either equation to find g.
Now, substitute –3 for j in either equation to find k.
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6-4 Elimination
Using Multiplication
The solution is (–1, –1).
40. 5j + 3k = −9
3j + 3k = −3
The solution is (–3, 2).
41. 2x − 4z = 6
x − 4z = −3
SOLUTION: Notice the coefficients for the k–terms are the same,
multiply equation 2 by –1 and add the equations to
find j .
SOLUTION: Notice the coefficients for the z–terms are the same,
so multiply equation 2 by –1 and add the equations to
find x.
Now, substitute –3 for j in either equation to find k.
Now, substitute 9 for x in either equation to find z.
The solution is (9, 3).
The solution is (–3, 2).
41. 2x − 4z = 6
x − 4z = −3
SOLUTION: Notice the coefficients for the z–terms are the same,
so multiply equation 2 by –1 and add the equations to
find x.
42. −5c − 3v = 9
5c + 2v = −6
SOLUTION: Notice the coefficients for the c–terms are the
opposite, so add the equations.
Now, substitute –3 for v in either equation to find c.
Now, substitute 9 for x in either equation to find z.
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6-4 Elimination
Using Multiplication
The solution is (9, 3).
42. −5c − 3v = 9
5c + 2v = −6
The solution is (0, –3).
43. 4b − 6n = −36
3b − 6n = −36
SOLUTION: Notice the coefficients for the c–terms are the
opposite, so add the equations.
SOLUTION: Notice the coefficients for the n–terms are the same,
so multiply equation 2 by –1 and add the equations to
find b.
Now, substitute –3 for v in either equation to find c.
Now, substitute 0 for b in either equation to find n.
The solution is (0, –3).
43. 4b − 6n = −36
3b − 6n = −36
SOLUTION: Notice the coefficients for the n–terms are the same,
so multiply equation 2 by –1 and add the equations to
find b.
The solution is (0, 6).
44. JOBS Brandy and Adriana work at an after–school
child care center. Together they cared for 32
children this week. Brandy cared for 0.6 times as
many children as Adriana. How many children did
each girl care for?
SOLUTION: Let b represent the number kids Brandy watched
and a represent the number of kids Adriana
watched.
b + a = 32
b = 0.6a
Now, substitute 0 for b in either equation to find n.
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Substitute 0.6a for b in the first equation to find the
value of a.
Page 17
Now, substitute 20 for a in either equation to find b.
The solution set is {m|m 13 and m −3}.
To graph the solution set, graph m 13 and graph m
−3. Then find the intersection.
6-4 Elimination
Using Multiplication
The solution is (0, 6).
44. JOBS Brandy and Adriana work at an after–school
child care center. Together they cared for 32
children this week. Brandy cared for 0.6 times as
many children as Adriana. How many children did
each girl care for?
SOLUTION: Let b represent the number kids Brandy watched
and a represent the number of kids Adriana
watched.
46. |q + 11| < 5
SOLUTION: and
The solution set is {q|q < −6 and q > −16}.
To graph the solution set, graph q < −6 and graph q >
−16. Then find the intersection.
b + a = 32
b = 0.6a
Substitute 0.6a for b in the first equation to find the
value of a.
47. |2w + 9| > 11
SOLUTION: The solution set is {w|w > 1 or w < −10}.
Notice that the graphs do not intersect. To graph the
solution set, graph w > 1 and graph w < −10. Then
find the union.
or
Now, substitute 20 for a in either equation to find b.
So, Brandy cared for 12 children and Adriana cared
for 20 children.
48. |2r + 1|
9
SOLUTION: or
Solve each inequality. Then graph the solution
set.
45. |m − 5| 8
SOLUTION: and
The solution set is {m|m 13 and m −3}.
To graph the solution set, graph m 13 and graph m
−3. Then find the intersection.
The solution set is {r|r 4 or r −5}.
Notice that the graphs do not intersect. To graph the
solution set, graph r 4 and graph r −5. Then find
the union.
Translate each sentence into a formula.
49. The area A of a triangle equals one half times the
base b times the height h.
46. |q + 11| < 5
SOLUTION: and
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The solution set is {q|q < −6 and q > −16}.
To graph the solution set, graph q < −6 and graph q >
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The area A of a triangle equals one–half
times the base b times the height h.
Page 18
A
equals
times
b
times
h
Notice that the graphs do not intersect. To graph the
solution set, graph r 4 and graph r −5. Then find
the union.
6-4 Elimination Using Multiplication
Translate each sentence into a formula.
49. The area A of a triangle equals one half times the
base b times the height h.
53. The area of a circle A equals the product of π and
the radius r squared.
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The area A of a triangle equals one–half
times the base b times the height h.
A
equals
times
b
times
h
A
•
=
b
•
the product of π and the radius r to the second
power multiplied by the height h.
V equals π times r squared times h
2
V
=
r
h
π
•
•
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The area of a circle A equals the product
of π and the radius r squared.
A
equals
times
r
squared
π
2
A
=
r
π
•
h
54. Acceleration A equals the increase in speed s divided
by time t in seconds.
50. The circumference C of a circle equals the product
of 2, π, and the radius r.
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The circumference C of a circle equals
the product of 2, π , and the radius r.
C
equals
2
times
times
r
π
C
=
2
r
•
π
•
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. Acceleration A equals speed s divided by
time t in seconds.
A
equals
s
divided by t
A
=
s
t
÷
51. The volume V of a rectangular box is the length
times the width w multiplied by the height h.
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The volume V of a rectangular box is the
length times the width w multiplied by the heigh
h.
V
equals
times
w
times
h
V
=
•
w
•
h
52. The volume of a cylinder V is the same as the
product of π and the radius r to the second power
multiplied by the height h.
SOLUTION: Rewrite the verbal sentence so it is easier to
translate. The volume V of a cylinder is the same as
the product of π and the radius r to the second
power multiplied by the height h.
V equals π times r squared times h
2
V
=
r
h
π
•
•
53. The area of a circle A equals the product of π and
the radius r squared.
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SOLUTION: Page 19