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, 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.
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Now, substitute 1 for y in either equation to find x.
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6-4 Elimination
Using
The solution is
(3, 2).Multiplication
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.
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.
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6-4 Elimination
Using Multiplication
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).
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.
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Now, substitute 0 for a in either equation to find b.
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6-4 Elimination
Using Multiplication
The solution is (–4, 1).
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).
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. eSolutions Manual - Powered by Cognero
with speed of speed of time
d
kayak
river (hours)
x
y
2
Page 4
d = rt
16 16 = 2(x + y)
6-4 Elimination
Using Multiplication
The solution is (0, 4).
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)
Notice that the coefficients of the y–terms are additive inverses, so add the equations.
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 ofPage
the 5
Hobbies and Recreation episodes lasted about 32 minutes. Each Soliloquies episodes lasted 42 minutes. To how many of each tag did Steve subscribe?
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SOLUTION: 6-4 Elimination
Using Multiplication
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?
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.
Steve subscribed to 2 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.
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6-4 Elimination
Using Multiplication
Steve subscribed to 2 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 (–1, 3).
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.
Now, substitute –2 for x in either equation to find y.
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6-4 Elimination
Using
Multiplication
The solution is
(–1, 3).
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.
Now, substitute –2 for x in either equation to find y.
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.
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6-4 Elimination
Using Multiplication
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).
10. 6x + y = −39
3x + 2y = −15
SOLUTION: Notice that if you multiply the first equation by –2, the coefficients of the y–terms are additive inverses.
Now, substitute –7 for x in either equation to find y.
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6-4 Elimination
Using Multiplication
The solution is (–3, 4).
10. 6x + y = −39
3x + 2y = −15
SOLUTION: Notice that if you multiply the first equation by –2, the coefficients of the y–terms are additive inverses.
Now, substitute –7 for x in either equation to find y.
The solution is (–7, 3).
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.
Now, substitute 3 for y in either equation to find x.
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6-4 Elimination
Using Multiplication
The solution is (–7, 3).
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.
Now, substitute 3 for y in either equation to find x.
The solution is (–2, 3).
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.
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6-4 Elimination
Using Multiplication
The solution is (–2, 3).
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.
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.
Now, substitute 5 for y in either equation to find x.
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6-4 Elimination
Using Multiplication
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.
Now, substitute 5 for y in either equation to find x.
The solution is (3, 5).
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.
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6-4 Elimination
Using Multiplication
The solution is (3, 5).
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 (2, –4).
15. 8x + 3y = −7
7x + 2y = −3
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.
Now, substitute 1 for x in either equation to find y.
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6-4 Elimination
Using Multiplication
The solution is (2, –4).
15. 8x + 3y = −7
7x + 2y = −3
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.
Now, substitute 1 for x in either equation to find y.
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
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equation to find x.
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6-4 Elimination
Using Multiplication
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.
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6-4 Elimination
Using Multiplication
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).
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.
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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.
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Notice that if you multiply the second equation by –3, the coefficients of the y–terms are additive inverses.
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6-4 Elimination
Using Multiplication
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.
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.
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Page 19
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.
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 find y.
So, he made 14 field goals and 7 extra point kicks.
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.
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Now, substitute 2.5 for x in either equation to find y.
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6-4 Elimination
Using Multiplication
So, he made 14 field goals and 7 extra point kicks.
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 2.5 for x in either equation to find y.
So, the solution is (2.5, 3.25).
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.
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Now, substitute 4.5 for x in either equation to find y.
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6-4 Elimination
Using
Multiplication
So, the solution
is (2.5,
3.25).
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.
Now, substitute 4.5 for x in either equation to find y.
So, the solution is (4.5, –1.5).
23. SOLUTION: Notice that if you multiply the second equation by –8, the coefficients of the y–terms are additive inverses.
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Now, substitute 3 for x in either equation to find y.
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6-4 Elimination
Using Multiplication
So, the solution is (4.5, –1.5).
23. SOLUTION: Notice that if you multiply the second equation by –8, the coefficients of the y–terms are additive inverses.
Now, substitute 3 for x in either equation to find y.
So, the solution is
.
24. eSolutions
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6-4 Elimination
Using
So, the solution
is Multiplication
.
24. SOLUTION: Notice that if you multiply the second equation by –12, the coefficients of the y–terms are additive inverses.
Now, substitute
for x in either equation to find y.
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So, the
solution
is by Cognero
.
Page 24
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
6-4 Elimination Using Multiplication
So, the solution is
.
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. 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
b. 90n + 120s = 1050
c. Notice that if you multiply the second equation by –3, the coefficients of the s–terms are additive inverses.
Now, substitute 5 for n in either equation to find s.
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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.
6-4 Elimination Using Multiplication
So, the solution is
.
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. 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
b. 90n + 120s = 1050
c. Notice that if you multiply the second equation by –3, the coefficients of the s–terms are additive inverses.
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.
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a. What are the coordinates of the vertex?
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6-4 Elimination Using Multiplication
(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 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.
Now, substitute 1 for y in either equation to find x.
The vertex is (4, 1).
b. Rewrite each equation in slope-intercept form and then graph on the same coordinate plane.
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Page 27
The vertex is (4, 1).
6-4 Elimination
Using Multiplication
b. Rewrite each equation in slope-intercept form and then graph on the same coordinate plane.
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.
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.
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Substitute 3 for y in either equation to find the value of x.
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6-4 Elimination Using Multiplication
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.
Substitute 3 for y in either equation to find the value of x.
So, the vertex is (0, 3).
Substitute 2 for x in either equation to find the value of y .
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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a. Define the variables, and write a system of linear equations from this situation.
b. Solve the system of equations, and explain what the solution represents.
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6-4 Elimination Using Multiplication
a. Define the variables, and write a system of linear equations from this situation.
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.
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.
The solution is (1.5, 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.
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the digits
is 14, so
x + y = 14. Since Mrs. Henderson switched the digits, she gave the students 10y + x points, when they actually earnedPage 30
10x + y points.The difference in points owed and points given is 36. 6-4 Elimination
Using Multiplication
The solution is (1.5, 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. 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,
as, 0by=Cognero
0. You can also notice that 12x - 30y = 84 is 6(2x - 5y = 14). The system of equations will have infinitely many solutions whenever
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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.
6-4 Elimination
Using Multiplication
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.
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 t-term 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.
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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: 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, 0Multiplication
= 0. You can also notice that 12x - 30y = 84 is 6(2x - 5y = 14). The system of equations will have infinitely many solutions whenever
6-4 Elimination
Using
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.
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 t-term 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
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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 t-term should not be eliminated. She needs to find
6-4 Elimination
Using
Multiplication
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
Substitute
for x in either equation and solve for 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.
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6-4 Elimination Using Multiplication
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?
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)
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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 Multiplication
reduced to 1 without turning other coefficients into fractions. Otherwise, elimination is more helpful because
6-4 Elimination
Using
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)
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?
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6-4 Elimination
Using Multiplication
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.
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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
6-4 Elimination
Using
Multiplication
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.
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
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6-4 Elimination Using Multiplication
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
.
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.
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Find the total number of outcomes = 4 + 8 + 2 + 5 + 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.
Now, substitute –1 for f 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.
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6-4 Elimination
Using
Multiplication
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 h in either equation to find g.
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 .
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6-4 Elimination
Using Multiplication
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 –3 for j in either equation to find k.
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.
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6-4 Elimination
Using Multiplication
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.
Now, substitute 9 for x in either equation to find z.
The solution is (9, 3).
42. −5c − 3v = 9
5c + 2v = −6
SOLUTION: Notice the coefficients for the c–terms are the opposite, so add the equations.
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6-4 Elimination
Using Multiplication
The solution is (9, 3).
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.
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.
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6-4 Elimination
Using Multiplication
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.
Now, substitute 0 for b in either equation to find n.
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
Substitute 0.6a for b in the first equation to find the value of a.
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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.
b + a = 32
b = 0.6a
Substitute 0.6a for b in the first equation to find the value of a.
Now, substitute 20 for a in either equation to find b.
So, Brandy cared for 12 children and Adriana cared for 20 children.
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
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46. |q + 11| < 5
SOLUTION: −3. Then find the intersection.
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6-4 Elimination
Using Multiplication
So, Brandy cared for 12 children and Adriana cared for 20 children.
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.
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.
47. |2w + 9| > 11
SOLUTION: or
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.
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48. |2r +Manual
SOLUTION: or
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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.
6-4 Elimination Using Multiplication
48. |2r + 1|
9
SOLUTION: or
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.
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
•
h
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
•
51. The volume V of a rectangular box is the length
times the width w multiplied by the height h.
SOLUTION: Rewrite
the- Powered
verbal sentence
eSolutions
Manual
by Cogneroso it is easier to translate. The volume V of a rectangular box is the length
V
equals
times
w
times
h
V
=
•
w
•
h
times the width w multiplied by the height h.
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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
=Using Multiplication
2
r
•
π
•
6-4 Elimination
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
V
equals
times
w
times
h
V
=
•
w
•
h
times the width w multiplied by the height 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
times
r
squared times
h
π
2
V
=
r
h
π
•
•
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 of a circle A equals the product of π and the radius r squared.
A
equals
times
r
squared
π
2
A
=
r
π
•
54. Acceleration A equals the increase in speed s divided by time t in seconds.
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
÷
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