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Study Guide and Review - Chapter 6
State whether each sentence is true or false. If false, replace the underlined term to make a true
sentence.
1. If a system has at least one solution, it is said to be consistent.
SOLUTION: If a system has at least one solution, it is said to be consistent. So, the statement is true.
2. If a consistent system has exactly two solution(s), it is said to be independent.
SOLUTION: The statement is false. If a consistent system has exactly one solution(s), it is said to be independent.
3. If a consistent system has an infinite number of solutions, it is said to be inconsistent.
SOLUTION: The statement is false. If a consistent system has an infinite number of solutions, it is said to be dependent.
4. If a system has no solution, it is said to be inconsistent.
SOLUTION: If a system has no solution, it is said to be inconsistent. So, the statement is true.
5. Substitution involves substituting an expression from one equation for a variable in the other.
SOLUTION: Substitution involves substituting an expression from one equation for a variable in the other. So, the statement is
true.
6. In some cases, dividing two equations in a system together will eliminate one of the variables. This process is called
elimination.
SOLUTION: The statement is false. In some cases, when adding or subtracting two equations in a system together will eliminate
one of the variables, this process is called elimination.
7. A set of two or more inequalities with the same variables is called a system of equations.
SOLUTION: The statement is false. A set of two or more inequalities with the same variables is called a system of inequalities.
8. When the graphs of the inequalities in a system of inequalities do not intersect, there are no solutions to the system.
SOLUTION: True
Graph each system and determine the number of solutions that it has. If it has one solution, name it.
9. x − y = 1
x +y = 5
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
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Page 1
The statement is false. A set of two or more inequalities with the same variables is called a system of inequalities.
8. When the graphs of the inequalities in a system of inequalities do not intersect, there are no solutions to the system.
SOLUTION: Study
Guide and Review - Chapter 6
True
Graph each system and determine the number of solutions that it has. If it has one solution, name it.
9. x − y = 1
x +y = 5
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
Equation 2:
Graph and locate the solution.
y=x−1
y = −x + 5
The graphs appear to intersect at the point (3, 2). You can check this by substituting 3 for x and 2 for y.
The solution is (3, 2).
10. y = 2x − 4
4x + y = 2
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 2:
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Page 2
Study
Guide
andisReview
The
solution
(3, 2). - Chapter 6
10. y = 2x − 4
4x + y = 2
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 2:
Graph and find the solution.
y = 2x − 4
y = −4x + 2
The graphs appear to intersect at the point (1, −2). You can check this by substituting 1 for x and −2 for y.
The solution is (1, −2).
11. 2x − 3y = −6
y = −3x + 2
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
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Page 3
Guide and Review - Chapter 6
Study
The solution is (1, −2).
11. 2x − 3y = −6
y = −3x + 2
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
Graph and find the solution.
y=
x+2
y = −3x + 2
The graphs appear to intersect at the point (0, 2). You can check this by substituting 0 for x and 2 for y.
The solution is (0, 2).
12. −3x + y = −3
y =x−3
SOLUTION: To graph the system, write both equations in slope-intercept form.
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Equation 1: Page 4
Guide and Review - Chapter 6
Study
The solution is (0, 2).
12. −3x + y = −3
y =x−3
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1: Graph and find the solution.
y = 3x − 3
y=x−3
The graphs appear to intersect at the point (0, −3). You can check this by substituting 0 for x and −3 for y.
The solution is (0, −3).
13. x + 2y = 6
3x + 6y = 8
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
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Page 5
Guide and Review - Chapter 6
Study
The solution is (0, −3).
13. x + 2y = 6
3x + 6y = 8
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
Equation 2:
Graph each equation.
y=
x+3
y=
x+
The lines have the same slope but different y-intercepts, so the lines are parallel. Since they do not intersect, there is
no solution of this system. The system is inconsistent.
eSolutions
14. 3x +Manual
y = 5 - Powered by Cognero
6x = 10 − 2y
Page 6
Study
Guide
The
lines and
haveReview
the same- Chapter
slope but 6different y-intercepts, so the lines are parallel. Since they do not intersect, there is
no solution of this system. The system is inconsistent.
14. 3x + y = 5
6x = 10 − 2y
SOLUTION: To graph the system, write both equations in slope-intercept form.
Equation 1:
Equation 2:
Graph the equations.
y = −3x + 5
y = −3x + 5
The two lines are identical, so there are an infinite number of solutions to the system. The system is dependent.
15. MAGIC NUMBERS Sean is trying to find two numbers with a sum of 14 and a difference of 4. Define two
variables, write a system of equations, and solve by graphing.
SOLUTION: Sample answer: Let x be one number and y be the other number.
x + y = 14
x − yManual
= 4 - Powered by Cognero
eSolutions
To graph the system, write both equations in slope-intercept form.
Page 7
Guide and Review - Chapter 6
Study
The two lines are identical, so there are an infinite number of solutions to the system. The system is dependent.
15. MAGIC NUMBERS Sean is trying to find two numbers with a sum of 14 and a difference of 4. Define two
variables, write a system of equations, and solve by graphing.
SOLUTION: Sample answer: Let x be one number and y be the other number.
x + y = 14
x −y = 4
To graph the system, write both equations in slope-intercept form.
Equation 1:
Equation 2:
Graph the equations and find the solution.
y = −x + 14
y=x−4
The graphs appear to intersect at the point (9, 5). So, the numbers 9 and 5 have a sum of 14 and a difference of 4.
Use substitution to solve each system of equations.
16. x + y = 3
x = 2y
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SOLUTION: x +y = 3
Page 8
Guide and Review - Chapter 6
Study
The graphs appear to intersect at the point (9, 5). So, the numbers 9 and 5 have a sum of 14 and a difference of 4.
Use substitution to solve each system of equations.
16. x + y = 3
x = 2y
SOLUTION: x +y = 3
x = 2y
Substitute 2y for x in the first equation.
Use the solution for y and either equation to find x.
x = 2y
x = 2(1)
x =2
The solution is (2, 1).
17. x + 3y = −28
y = −5x
SOLUTION: x + 3y = −28
y = −5x
Substitute −5x for y in the first equation.
Use the solution for x and either equation to find y.
The solution is (2, −10).
18. 3x + 2y = 16
x = 3y − 2
SOLUTION: 3x +Manual
2y = 16
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x = 3y − 2
Page 9
Guide and Review - Chapter 6
Study
The solution is (2, −10).
18. 3x + 2y = 16
x = 3y − 2
SOLUTION: 3x + 2y = 16
x = 3y − 2
Substitute 3y − 2 for x in the first equation.
Use the solution for y and either equation to find x.
The solution is (4, 2).
19. x − y = 8
y = −3x
SOLUTION: x −y = 8
y = −3x
Substitute −3x for y in the first equation.
Use the solution for x and either equation to find y.
The solution is (2, −6).
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20. y = 5x − 3
x + 2y = 27
Page 10
Guide and Review - Chapter 6
Study
The solution is (4, 2).
19. x − y = 8
y = −3x
SOLUTION: x −y = 8
y = −3x
Substitute −3x for y in the first equation.
Use the solution for x and either equation to find y.
The solution is (2, −6).
20. y = 5x − 3
x + 2y = 27
SOLUTION: y = 5x − 3
x + 2y = 27
Substitute 5x − 3 for y in the second equation.
Use the solution for x and either equation to find y.
The solution is (3, 12).
21. x + 3y
= 9 - Powered by Cognero
eSolutions
Manual
x +y = 1
SOLUTION: Page 11
Study
Guide and Review - Chapter 6
The solution is (3, 12).
21. x + 3y = 9
x +y = 1
SOLUTION: x + 3y = 9
x +y = 1
First, solve the second equation for y to get y = −x +1. Then substitute −x + 1 for y in the first equation.
Use the solution for x and either equation to find y.
The solution is (−3, 4).
22. GEOMETRY The perimeter of a rectangle is 48 inches. The length is 6 inches greater than the width. Define the
variables, and write equations to represent this situation. Solve the system by using substitution.
SOLUTION: Sample answer: Let w be the width and be the length.
2 + 2w = 48
=w+6
Substitute w + 6 for in the first equation.
Use the solution for w and either equation to find .
=w+6
=9+6
= 15
The solution is (9, 15).
Use Manual
elimination
each
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Cognero
23. x + y = 13
x −y = 5
system of equations.
Page 12
=w+6
=9+6
= 15
Study
Guide and Review - Chapter 6
The solution is (9, 15).
Use elimination to solve each system of equations.
23. x + y = 13
x −y = 5
SOLUTION: Because y and −y have opposite coefficients, add the equations.
Now, substitute 9 for x in either equation to find y.
The solution is (9, 4).
24. −3x + 4y = 21
3x + 3y = 14
SOLUTION: Because −3x and 3x have opposite coefficients, add the equations.
Now, substitute 5 for y in either equation to find x.
The solution is
.
25. x + 4y = −4
x + 10y = −16
SOLUTION: eSolutions
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Because x and x have the same coefficients, multiply equation 1 by –1 so the x's are additive inverses. Then add the
equations.
Study
Guide
andisReview - .Chapter 6
The
solution
25. x + 4y = −4
x + 10y = −16
SOLUTION: Because x and x have the same coefficients, multiply equation 1 by –1 so the x's are additive inverses. Then add the
equations.
Now, substitute −2 for y in either equation to find x.
The solution is (4, −2).
26. 2x + y = −5
x −y = 2
SOLUTION: Because y and −y have opposite coefficients, add the equations.
Now, substitute −1 for x in either equation to find y.
The solution is (−1, −3).
27. 6x + y = 9
−6x + 3y = 15
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SOLUTION: Page 14
Guide and Review - Chapter 6
Study
The solution is (−1, −3).
27. 6x + y = 9
−6x + 3y = 15
SOLUTION: Because 6x and −6x have opposite coefficients, add the equations.
Now, substitute 6 for y in either equation to find x.
The solution is
.
28. x − 4y = 2
3x + 4y = 38
SOLUTION: Because −4y and 4y have opposite coefficients, add the equations.
Now, substitute 10 for x in either equation to find y.
The solution is (10, 2).
29. 2x + 2y = 4
2x − 8y = −46
SOLUTION: Because 2x and 2x have the same coefficients, you need to multiply equation 2 by -1 so they are additive inverses.
Then add the equations.
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Study
Guide and Review - Chapter 6
The solution is (10, 2).
29. 2x + 2y = 4
2x − 8y = −46
SOLUTION: Because 2x and 2x have the same coefficients, you need to multiply equation 2 by -1 so they are additive inverses.
Then add the equations.
Now, substitute 5 for y in either equation to find x.
The solution is (−3, 5).
30. 3x + 2y = 8
x + 2y = 2
SOLUTION: Because 2y and 2y have same coefficients, multiply equation 2 by –1 so the terms are additive inverses. Then add
the equations.
Now, substitute 3 for x in either equation to find y.
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Study
Guide and Review - Chapter 6
The solution is (−3, 5).
30. 3x + 2y = 8
x + 2y = 2
SOLUTION: Because 2y and 2y have same coefficients, multiply equation 2 by –1 so the terms are additive inverses. Then add
the equations.
Now, substitute 3 for x in either equation to find y.
The solution is
.
31. BASEBALL CARDS Cristiano bought 24 baseball cards for $50. One type cost $1 per card, and the other cost $3
per card. Define the variables, and write equations to find the number of each type of card he bought. Solve by using
elimination.
SOLUTION: Sample answer: Let f be the first type of card and let c be the second type of card.
f + c = 24 (total number of cards)
f + 3c = 50 (cost of the cards)
Because f and f have the same coefficients, multiply equation 2 by –1 and then add the equations.
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by Cognero
Now,
substitute
13 for
c in either
equation to find f .
Page 17
Study
Guide
andisReview - .Chapter 6
The
solution
31. BASEBALL CARDS Cristiano bought 24 baseball cards for $50. One type cost $1 per card, and the other cost $3
per card. Define the variables, and write equations to find the number of each type of card he bought. Solve by using
elimination.
SOLUTION: Sample answer: Let f be the first type of card and let c be the second type of card.
f + c = 24 (total number of cards)
f + 3c = 50 (cost of the cards)
Because f and f have the same coefficients, multiply equation 2 by –1 and then add the equations.
Now, substitute 13 for c in either equation to find f .
The solution is (11, 13).
Cristiano bought 11 $1 cards and 13 $3 cards.
Use elimination to solve each system of equations.
32. x + y = 4
−2x + 3y = 7
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 (1, 3).
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33. x − y = −2
2x + 4y = 38
Page 18
Study
Guide
andisReview
The
solution
(11, 13).- Chapter 6
Cristiano bought 11 $1 cards and 13 $3 cards.
Use elimination to solve each system of equations.
32. x + y = 4
−2x + 3y = 7
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 (1, 3).
33. x − y = −2
2x + 4y = 38
SOLUTION: Notice that if you multiply the first equation by 4, the coefficients of the y-terms are additive inverses.
Now, substitute 5 for x in either equation to find y.
The solution is (5, 7).
34. 3x + 4y = 1
5x + 2y = 11
SOLUTION: Notice that if you multiply the second equation by −2, the coefficients of the y-terms are additive inverses.
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Study
Guide and Review - Chapter 6
The solution is (5, 7).
34. 3x + 4y = 1
5x + 2y = 11
SOLUTION: Notice that if you multiply the second equation by −2, 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).
35. −9x + 3y = −3
3x − 2y = −4
SOLUTION: Notice that if you multiply the second equation by 3, the coefficients of the x-terms are additive inverses.
Now, substitute 5 for y in either equation to find x.
The solution is (2, 5).
36. 8x − 3y = −35
3x + 4y = 33
SOLUTION: Notice that if you multiply the first equation by 4 and the second equation by 3, the coefficients of the y-terms are
additive inverses.
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Study
Guide and Review - Chapter 6
The solution is (2, 5).
36. 8x − 3y = −35
3x + 4y = 33
SOLUTION: Notice that if you multiply the first equation by 4 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, 9).
37. 2x + 9y = 3
5x + 4y = 26
SOLUTION: Notice that if you multiply the first equation by 5 and the second equation by −2, the coefficients of x-terms are
additive inverses.
Now, substitute −1 for y in either equation to find x.
The solution is (6, −1).
38. −7x + 3y = 12
2x − 8y = −32
SOLUTION: Notice
that- ifPowered
you multiply
the first equation by 2 and the second equation by 7, the coefficients of the x-terms are
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by Cognero
Page 21
additive inverses.
Guide and Review - Chapter 6
Study
The solution is (6, −1).
38. −7x + 3y = 12
2x − 8y = −32
SOLUTION: Notice that if you multiply the first equation by 2 and the second equation by 7, the coefficients of the x-terms are
additive inverses.
Now, substitute 4 for y in either equation to find x.
The solution is (0, 4).
39. 8x − 5y = 18
6x + 6y = −6
SOLUTION: Notice that if you multiply the first equation by 6 and the second equation by 5, the coefficients of the y-terms are
additive inverses.
Now, substitute 1 for x in either equation to find y.
The solution is (1, −2).
40. BAKE SALE On the first day, a total of 40 items were sold for $356. Define the variables, and write a system of
equations to find the number of cakes and pies sold. Solve by using elimination.
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Guide and Review - Chapter 6
Study
The solution is (1, −2).
40. BAKE SALE On the first day, a total of 40 items were sold for $356. Define the variables, and write a system of
equations to find the number of cakes and pies sold. Solve by using elimination.
SOLUTION: Let c represent the cakes and let p represent the pies.
8c + 10p = 356
c + p = 40
Notice that if you multiply the second equation by −8, the coefficients of the c-terms are additive inverses.
Now, substitute 18 for p in either equation to find c.
The solution is (22, 18).
The Monarch Band Booster sold 22 cakes and 18 pies.
Determine the best method to solve each system of equations. Then solve the system.
41. y = x − 8
y = −3x
SOLUTION: Because both equations are solved for one of the variables, substitution is the best method.
Substitute −3x for y in the first equation.
Substitute 2 for x in either equation to find y.
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Page 23
The
solution
(22, 18).- Chapter 6
Study
Guide
andisReview
The Monarch Band Booster sold 22 cakes and 18 pies.
Determine the best method to solve each system of equations. Then solve the system.
41. y = x − 8
y = −3x
SOLUTION: Because both equations are solved for one of the variables, substitution is the best method.
Substitute −3x for y in the first equation.
Substitute 2 for x in either equation to find y.
The solution is (2, −6).
42. y = −x
y = 2x
SOLUTION: Because both equations are solved for one of the variables, substitution is the best method.
Substitute 2x for y in the first equation.
Substitute 0 for x in either equation to find y.
The solution is (0, 0).
43. x + 3y = 12
x = −6y
SOLUTION: Because one of the equations is solved for one of the variables, substitution is the best method.
Substitute −6y for x in the first equation.
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Study
Guide and Review - Chapter 6
The solution is (0, 0).
43. x + 3y = 12
x = −6y
SOLUTION: Because one of the equations is solved for one of the variables, substitution is the best method.
Substitute −6y for x in the first equation.
Substitute −4 for y in either equation to find x.
The solution is (24, −4).
44. x + y = 10
x − y = 18
SOLUTION: Because y and −y have opposite coefficients, elimination using addition is the best method.
Now, substitute 14 for x in either equation to find y.
The solution is (14, −4).
45. 3x + 2y = −4
5x + 2y = −8
SOLUTION: Because 2y and 2y have the same coefficient, elimination using addition is the best method. Multiply equation 2 by –
1. eSolutions Manual - Powered by Cognero
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Guide and Review - Chapter 6
Study
The solution is (14, −4).
45. 3x + 2y = −4
5x + 2y = −8
SOLUTION: Because 2y and 2y have the same coefficient, elimination using addition is the best method. Multiply equation 2 by –
1. Now, substitute −2 for x in either equation to find y.
The solution is (−2, 1).
46. 6x + 5y = 9
−2x + 4y = 14
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the second equation by 3, the coefficients of the x-terms are additive inverses.
Now, substitute 3 for y in either equation to find x.
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3).Cognero
47. 3x + 4y = 26
Page 26
Study
Guide and Review - Chapter 6
The solution is (−2, 1).
46. 6x + 5y = 9
−2x + 4y = 14
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the second 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).
47. 3x + 4y = 26
2x + 3y = 19
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the first equation by 2 and the second equation by −3, the coefficients of the x-terms are
additive inverses.
Now, substitute 5 for y in either equation to find x.
The solution is (2, 5).
48. 11x − 6y = 3
5x − 8y = −25
SOLUTION: Page 27
Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the first equation by 4 and the second equation by −3, the coefficients of the y-terms are
eSolutions Manual - Powered by Cognero
Guide and Review - Chapter 6
Study
The solution is (2, 5).
48. 11x − 6y = 3
5x − 8y = −25
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the first equation by 4 and the second 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, 5).
49. COINS Tionna has 25 coins in her piggy bank with a value of $4. The coins are either dimes or quarters. Define
the variables, and write a system of equations to determine the number of dimes and quarters. Then solve the system
using the best method for the situation.
SOLUTION: Sample answer: Let d represent the number of dimes and let q represent the number of quarters. Use the fact that
the value of a dime is $0.10 and the value of a quarter is $0.25 to write the equation for the value.
d + q = 25
0.10d + 0.25q = 4
Because the coefficients of d and q in the first equation are 1, the best method is substitution. Solve the first equation
for q.
q = −d + 25
Substitute −d + 25 for q in the second equation.
Substitute 15 for d in either equation to find q.
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The solution is (15, 10).
Page 28
Study
Guide and Review - Chapter 6
The solution is (3, 5).
49. COINS Tionna has 25 coins in her piggy bank with a value of $4. The coins are either dimes or quarters. Define
the variables, and write a system of equations to determine the number of dimes and quarters. Then solve the system
using the best method for the situation.
SOLUTION: Sample answer: Let d represent the number of dimes and let q represent the number of quarters. Use the fact that
the value of a dime is $0.10 and the value of a quarter is $0.25 to write the equation for the value.
d + q = 25
0.10d + 0.25q = 4
Because the coefficients of d and q in the first equation are 1, the best method is substitution. Solve the first equation
for q.
q = −d + 25
Substitute −d + 25 for q in the second equation.
Substitute 15 for d in either equation to find q.
The solution is (15, 10).
So, Tionna has 15 dimes and 10 quarters in her piggy bank.
50. FAIR At a county fair, the cost for 4 slices of pizza and 2 orders of French fries is $21.00. The cost of 2 slices of
pizza and 3 orders of French fries is $16.50. To find out how much a single slice of pizza and an order of French
fries costs, define the variables and write a system of equations to represent the situation. Determine the best
method to solve the system of equations. Then solve the system.
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the second equation by -2, the coefficients of the p -terms are additive inverses.
Now, substitute 3 for f in either equation to find p .
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Page 29
The
solution
(15, 10).- Chapter 6
Study
Guide
andisReview
So, Tionna has 15 dimes and 10 quarters in her piggy bank.
50. FAIR At a county fair, the cost for 4 slices of pizza and 2 orders of French fries is $21.00. The cost of 2 slices of
pizza and 3 orders of French fries is $16.50. To find out how much a single slice of pizza and an order of French
fries costs, define the variables and write a system of equations to represent the situation. Determine the best
method to solve the system of equations. Then solve the system.
SOLUTION: Because none of the coefficients are 1 or −1, elimination using multiplication is the best method.
Notice that if you multiply the second equation by -2, the coefficients of the p -terms are additive inverses.
Now, substitute 3 for f in either equation to find p .
pizza: $3.75; French fries: $3
Solve each system of inequalities by graphing.
51. x > 3
y <x+2
SOLUTION: Graph each inequality. The graph of x > 3 is dashed and is not included in the graph of the solution. The graph of y < x + 2 is also dashed and is not included in the graph of the solution. eSolutions Manual - Powered by Cognero
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The solution of the system is the set of ordered pairs in the intersection of the graphs of x > 3 and y < x + 2. Overlay
the graphs and locate the green region. This is the intersection.
Study
Guide and Review - Chapter 6
pizza: $3.75; French fries: $3
Solve each system of inequalities by graphing.
51. x > 3
y <x+2
SOLUTION: Graph each inequality. The graph of x > 3 is dashed and is not included in the graph of the solution. The graph of y < x + 2 is also dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of x > 3 and y < x + 2. Overlay
the graphs and locate the green region. This is the intersection.
The solution region is shaded in the graph below.
52. y ≤ 5
y >x−4
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SOLUTION: Graph each inequality. Page 31
Study Guide and Review - Chapter 6
52. y ≤ 5
y >x−4
SOLUTION: Graph each inequality. The graph of y ≤ 5 is solid and is included in the graph of the solution. The graph of y > x − 4 is dashed and is not included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≤ 5 and y > x − 4. Overlay
the graphs and locate the green region. This is the intersection.
The solution region is shaded in the graph below.
53. y < 3x − 1
y ≥ −2x + 4
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SOLUTION: Graph each inequality. Page 32
Study Guide and Review - Chapter 6
53. y < 3x − 1
y ≥ −2x + 4
SOLUTION: Graph each inequality. The graph of y < 3x − 1 is dashed and is not included in the graph of the solution. The graph of y ≥ −2x + 4 is solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y < 3x − 1 and y ≥ −2x + 4.
Overlay the graphs and locate the green region. This is the intersection.
The solution region is shaded in the graph below.
54. y ≤ −x − 3
y ≥ 3x − 2
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SOLUTION: Graph each inequality.
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Study Guide and Review - Chapter 6
54. y ≤ −x − 3
y ≥ 3x − 2
SOLUTION: Graph each inequality.
The graph of y ≤ −x − 3 is solid and is included in the graph of the solution. The graph of y ≥ 3x − 2 is also solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of y ≤ −x − 3 and y ≥ 3x − 2.
Overlay the graphs and locate the green region. This is the intersection.
The solution region is shaded in the graph below.
55. JOBS Kishi makes $7 an hour working at the grocery store and $10 an hour delivering newspapers. She cannot
work more than 20 hours per week. Graph two inequalities that Kishi can use to determine how many hours she
needs to work at each job if she wants to earn at least $90 per week.
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SOLUTION: Let g represent the number of hours Kishi works at the grocery store. Let n represent the number of hours she
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Study Guide and Review - Chapter 6
55. JOBS Kishi makes $7 an hour working at the grocery store and $10 an hour delivering newspapers. She cannot
work more than 20 hours per week. Graph two inequalities that Kishi can use to determine how many hours she
needs to work at each job if she wants to earn at least $90 per week.
SOLUTION: Let g represent the number of hours Kishi works at the grocery store. Let n represent the number of hours she
spends delivering newspapers. Write a system of inequalities.
7g +10n ≥ 90
g + n ≤ 20
Graph each inequality. The graph of 7g +10n ≥ 90 is solid and is included in the graph of the solution. The graph of g + n ≤ 20 is also solid and is included in the graph of the solution. The solution of the system is the set of ordered pairs in the intersection of the graphs of 7g +10n ≥ 90 and g + n ≤
20. Overlay the graphs and locate the green region. This is the intersection.
The Manual
solution
region by
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in the graph below.
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Study Guide and Review - Chapter 6
The solution region is shaded in the graph below.
eSolutions Manual - Powered by Cognero
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