Download Equation 2 - Defiance City Schools

 The solution is (3, 2).
Study Guide and Review - Chapter 6
ANSWER: one; (3, 2)
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 slopeintercept form.
Equation 1:
10. y = 2x − 4
4x + y = 2
Equation 2:
Graph and locate the solution.
y=x−1
y = −x + 5
SOLUTION: To graph the system, write both equations in slopeintercept form.
Equation 2:
Graph and find the solution.
y = 2x − 4
y = −4x + 2
The graphs appear to intersect at the point (3, 2).
You can check this by substituting 3 for x and 2 for
y.
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 (3, 2).
ANSWER: one; (3, 2)
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The solution is (1, −2).
ANSWER: one; (1, −2)
Page 1
Study Guide and Review - Chapter 6
The solution is (1, −2).
ANSWER: one; (1, −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).
ANSWER: one; (0, 2)
11. 2x − 3y = −6
y = −3x + 2
SOLUTION: To graph the system, write both equations in slopeintercept form.
Equation 1:
12. −3x + y = −3
y =x−3
SOLUTION: To graph the system, write both equations in slopeintercept form.
Graph and find the solution.
y=
Equation 1: x+2
y = −3x + 2
Graph and find the solution.
y = 3x − 3
y=x−3
The graphs appear to intersect at the point (0, 2).
You can check this by substituting 0 for x and 2 for
y.
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Page 2
The graphs appear to intersect at the point (0, −3).
Graph and find the solution.
y =Guide
3x − 3and Review - Chapter 6
Study
y=x−3
SOLUTION: To graph the system, write both equations in slopeintercept form.
Equation 1:
Equation 2:
The graphs appear to intersect at the point (0, −3).
You can check this by substituting 0 for x and −3 for
y.
Graph each equation.
y=
x+3
y=
x+
The solution is (0, −3).
ANSWER: one; (0, −3)
13. x + 2y = 6
3x + 6y = 8
SOLUTION: To graph the system, write both equations in slopeintercept form.
The lines have the same slope but different yintercepts, so the lines are parallel. Since they do not
intersect, there is no solution of this system. The
system is inconsistent.
ANSWER: no solution
Equation 1:
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Equation 2:
Page 3
14. 3x + y = 5
6x = 10 − 2y
intercepts, so the lines are parallel. Since they do not
intersect, there is no solution of this system. The
system is inconsistent.
Study Guide and Review - Chapter 6
ANSWER: no solution
The two lines are identical, so there are an infinite
number of solutions to the system. The system is
dependent.
ANSWER: infinitely many solutions
14. 3x + y = 5
6x = 10 − 2y
SOLUTION: To graph the system, write both equations in slopeintercept form.
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.
Equation 2:
x + y = 14
x −y = 4
Equation 1:
To graph the system, write both equations in slopeintercept form.
Graph the equations.
y = −3x + 5
y = −3x + 5
Equation 1:
Equation 2:
Graph the equations and find the solution.
The two lines are identical, so there are an infinite
number of solutions to the system. The system is
dependent.
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ANSWER: infinitely many solutions
y = −x + 14
y=x−4
Page 4
Graph the equations and find the solution.
y =Guide
−x + 14
Study
and Review - Chapter 6
y=x−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.
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.
ANSWER: Sample answer: Let x be one number and y the other
number; x + y = 14; x − y = 4; 9 and 5
Use the solution for y and either equation to find x.
x = 2y
x = 2(1)
x =2
The solution is (2, 1).
ANSWER: (2, 1)
17. x + 3y = −28
y = −5x
Use substitution to solve each system of
equations.
16. x + y = 3
x = 2y
SOLUTION: x + 3y = −28
y = −5x
Substitute −5x for y in the first equation.
SOLUTION: x +y = 3
x = 2y
Substitute 2y for x in the first equation.
Use the solution for x and either equation to find y.
Use the solution for y and either equation to find x.
x = 2y
x = 2(1)
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x = 2Manual - Powered by Cognero
The solution is (2, 1).
The solution is (2, −10).
ANSWER: (2, −10)
Page 5
The solution is (2, 1).
ANSWER: Study
and Review - Chapter 6
(2,Guide
1)
17. x + 3y = −28
y = −5x
SOLUTION: x + 3y = −28
y = −5x
Substitute −5x for y in the first equation.
The solution is (2, −10).
ANSWER: (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 x and either equation to find y.
Use the solution for y and either equation to find x.
The solution is (2, −10).
ANSWER: (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.
The solution is (4, 2).
ANSWER: (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 (4, 2).
The solution is (2, −6).
ANSWER: ANSWER: Use the solution for y and either equation to find x.
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Page 6
The solution is (4, 2).
ANSWER: Study
Guide and Review - Chapter 6
(4, 2)
19. x − y = 8
y = −3x
The solution is (2, −6).
ANSWER: (2, −6)
20. y = 5x − 3
x + 2y = 27
SOLUTION: x −y = 8
y = −3x
Substitute −3x for y in the first equation.
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.
Use the solution for x and either equation to find y.
The solution is (2, −6).
ANSWER: (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.
The solution is (3, 12).
ANSWER: (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, 12).
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ANSWER: (3, 12)
Use the solution for x and either equation to find y.
The solution is (−3, 4).
ANSWER: (−3, 4)
Page 7
The solution is (3, 12).
ANSWER: Study
Guide and Review - Chapter 6
(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.
The solution is (−3, 4).
ANSWER: (−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 x and either equation to find y.
Use the solution for w and either equation to find .
The solution is (−3, 4).
ANSWER: (−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.
=w+6
=9+6
= 15
The solution is (9, 15).
ANSWER: Sample answer: Let w be the width and l be the
length; 2l + 2w = 48, l = w + 6; 9 is the width and 15
is the length.
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.
Use the solution for w and either equation to find .
=w+6
=
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eSolutions9Manual
= 15
The solution is (9, 15).
Now, substitute 9 for x in either equation to find y.
Page 8
ANSWER: Sample answer: Let w be the width and l be the
Study
Guide
length;
2l and
+ 2wReview
= 48, l =- Chapter
w + 6; 9 is6the width and 15
is the length.
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.
The solution is (9, 4).
ANSWER: (9, 4)
24. −3x + 4y = 21
3x + 3y = 14
SOLUTION: Because −3x and 3x have opposite coefficients, add
the equations.
Now, substitute 9 for x in either equation to find y.
Now, substitute 5 for y in either equation to find x.
The solution is (9, 4).
ANSWER: (9, 4)
The solution is
24. −3x + 4y = 21
3x + 3y = 14
.
ANSWER: SOLUTION: Because −3x and 3x have opposite coefficients, add
the equations.
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 5 for y in either equation to find x.
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The solution is
ANSWER: .
Now, substitute −2 for y in either equation to find x.
Page 9
ANSWER: Study Guide and Review - Chapter 6
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.
The solution is (4, −2).
ANSWER: (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.
Now, substitute −2 for y in either equation to find x.
The solution is (−1, −3).
ANSWER: (−1, −3)
The solution is (4, −2).
ANSWER: (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.
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
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The solution is (−1, −3).
ANSWER: .
Page 10
The solution is (−1, −3).
ANSWER: ANSWER: Study
Guide and Review - Chapter 6
(−1, −3)
27. 6x + y = 9
−6x + 3y = 15
28. x − 4y = 2
3x + 4y = 38
SOLUTION: Because 6x and −6x have opposite coefficients, add
the equations.
SOLUTION: Because −4y and 4y have opposite coefficients, add
the equations.
Now, substitute 6 for y in either equation to find x.
Now, substitute 10 for x in either equation to find y.
The solution is (10, 2).
The solution is
.
ANSWER: 28. x − 4y = 2
3x + 4y = 38
SOLUTION: Because −4y and 4y have opposite coefficients, add
the equations.
ANSWER: (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 10 for x in either equation to find y.
Now, substitute 5 for y in either equation to find x.
The solution is (10, 2).
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ANSWER: (10, 2)
Page 11
The solution is (−3, 5).
The solution is (10, 2).
The solution is (−3, 5).
ANSWER: Study
Guide and Review - Chapter 6
(10, 2)
29. 2x + 2y = 4
2x − 8y = −46
ANSWER: (−3, 5)
30. 3x + 2y = 8
x + 2y = 2
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.
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 5 for y in either equation to find x.
Now, substitute 3 for x in either equation to find y.
The solution is (−3, 5).
The solution is
ANSWER: (−3, 5)
.
ANSWER: 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.
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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Manual
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Now,
substitute
3 for
in either
Page 12
equation to find y.
ANSWER: ANSWER: Study Guide and Review - Chapter 6
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.
Sample answer: Let f be the number of the first type of
card, and let c be the number of the second type of
card; f + c = 24, f + 3c = 50; 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).
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.
ANSWER: (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.
ANSWER: Sample answer: Let f be the number of the first type of
card, and let c be the number of the second type of
card; f + c = 24, f + 3c = 50; 11 $1 cards and 13 $3
cards.
Now, substitute 5 for x in either equation to find y.
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.
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The solution is (5, 7).
ANSWER: (5, 7)
34. 3x + 4y = 1
5x + 2y = 11
Page 13
SOLUTION: Notice that if you multiply the second equation by −2,
The solution is (5, 7).
ANSWER: Study
Guide and Review - Chapter 6
(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).
ANSWER: (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).
ANSWER: (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 yterms are additive inverses.
Now, substitute −1 for x in either equation to find y.
The solution is (−1, 9).
ANSWER: (−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 xterms are additive inverses.
Now, substitute −1 for y in either equation to find x.
The solution is (2, 5).
ANSWER: (2, 5)
36. 8x − 3y = −35
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3x + 4y = 33
SOLUTION: The solution is (6, −1).
ANSWER: (6, −1)
38. −7x + 3y = 12
2x − 8y = −32
Page 14
The solution is (−1, 9).
ANSWER: Study
Guide and Review - Chapter 6
(−1, 9)
37. 2x + 9y = 3
5x + 4y = 26
The solution is (6, −1).
ANSWER: (6, −1)
38. −7x + 3y = 12
2x − 8y = −32
SOLUTION: Notice that if you multiply the first equation by 5 and
the second equation by −2, the coefficients of xterms are additive inverses.
SOLUTION: Notice that if you multiply the first equation by 2 and
the second equation by 7, the coefficients of the xterms are additive inverses.
Now, substitute −1 for y in either equation to find x.
The solution is (6, −1).
ANSWER: (6, −1)
ANSWER: (0, 4)
38. −7x + 3y = 12
2x − 8y = −32
39. 8x − 5y = 18
6x + 6y = −6
Now, substitute 4 for y in either equation to find x.
The solution is (0, 4).
SOLUTION: Notice that if you multiply the first equation by 2 and
the second equation by 7, the coefficients of the xterms are additive inverses.
SOLUTION: Notice that if you multiply the first equation by 6 and
the second equation by 5, the coefficients of the yterms are additive inverses.
Now, substitute 4 for y in either equation to find x.
Now, substitute 1 for x in either equation to find y.
The solution is (0, 4).
The solution is (1, −2).
ANSWER: (0, 4)
ANSWER: (1, −2)
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39. 8x − 5y = 18
6x + 6y = −6
Page 15
40. BAKE SALE On the first day, a total of 40 items
were sold for $356. Define the variables, and write a
The solution is (0, 4).
The solution is (1, −2).
ANSWER: Study
Guide and Review - Chapter 6
(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 yterms are additive inverses.
ANSWER: (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.
Now, substitute 1 for x in either equation to find y.
The solution is (1, −2).
ANSWER: (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.
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.
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ANSWER: Sample answer: Let c represent the number of the
cakes, and let p represent the number of pies; 8c +
10p = 356, p + c = 40; 22 cakes, 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.
Page 16
Substitute −3x for y in the first equation.
ANSWER: Sample answer: Let c represent the number of the
Study
Guide
- Chapter
6 of pies; 8c +
cakes,
andand
let pReview
represent
the number
10p = 356, p + c = 40; 22 cakes, 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.
ANSWER: Subs.;
(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).
The solution is (2, −6).
ANSWER: Subs.;
(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.
ANSWER: Subs.;
(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.
Substitute 0 for x in either equation to find y.
The solution is (24, −4).
The solution is (0, 0).
ANSWER: Subs.;
(0, 0)
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ANSWER: Subs.;
(24, −4)
44. x + y = 10
x − y = 18
Page 17
The solution is (0, 0).
ANSWER: Subs.;
Study
Guide and Review - Chapter 6
(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).
ANSWER: Subs.;
(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 (24, −4).
The solution is (14, −4).
ANSWER: Subs.;
(24, −4)
ANSWER: Elim (+); (14, −4)
44. x + y = 10
x − y = 18
SOLUTION: Because y and −y have opposite coefficients,
elimination using addition is the best method.
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 14 for x in either equation to find y.
Now, substitute −2 for x in either equation to find y.
The solution is (14, −4).
ANSWER: Elim (+); (14, −4)
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45. 3x + 2y = −4
5x + 2y = −8
Page 18
The solution is (14, −4).
ANSWER: Elim (+); Study
Guide and Review - Chapter 6
(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. The solution is (−2, 1).
ANSWER: Elim (–);
(−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.
Now, substitute −2 for x in either equation to find y.
The solution is (−1, 3).
ANSWER: Elim (×);
(−1, 3)
The solution is (−2, 1).
ANSWER: Elim (–);
(−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.
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 xterms are additive inverses.
Now, substitute 5 for y in either equation to find x.
Now, substitute 3 for y in either equation to find x.
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The solution is (2, 5).
ANSWER: Page 19
The solution is (−1, 3).
ANSWER: Study
Guide and Review - Chapter 6
Elim (×);
(−1, 3)
47. 3x + 4y = 26
2x + 3y = 19
The solution is (2, 5).
ANSWER: Elim (×);
(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 2 and
the second equation by −3, the coefficients of the xterms are additive inverses.
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 yterms are additive inverses.
Now, substitute 3 for x in either equation to find y.
Now, substitute 5 for y in either equation to find x.
The solution is (2, 5).
The solution is (3, 5).
ANSWER: Elim (×);
(2, 5)
ANSWER: Elim (×);
(3, 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 yterms are additive inverses.
Now, substitute 3 for x in either equation to find y.
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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.
Page 20
The solution is (3, 5).
ANSWER: Study
Guide and Review - Chapter 6
Elim (×);
(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.
ANSWER: Sample answer: Let d represent the number of dimes
and let q represent the number of quarters; d + q =
25, 0.10d + 0.25q = 4; 15 dimes, 10 quarters
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 .
Substitute 15 for d in either equation to find q.
pizza: $3.75; French fries: $3
The solution is (15, 10).
So, Tionna has 15 dimes and 10 quarters in her piggy
bank.
ANSWER: Sample answer: Let d represent the number of dimes
and let q represent the number of quarters; d + q =
25, 0.10d + 0.25q = 4; 15 dimes, 10 quarters
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
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method to solve the system of equations. Then solve
the system.
ANSWER: Let p represent the cost of a slice of pizza and t
represent the cost of an order of French fries; 4p +
2f = 21, 2p + 3f = 16.5; Sample answer: elimination;
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. Page 21
y <x+2
SOLUTION: Graph each inequality. Study
Guide and Review - Chapter 6
The graph of x > 3 is dashed and is not included in
the graph of the solution. ANSWER: The graph of y < x + 2 is also dashed and is not
included in the graph of the solution. 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 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 graph of y > x − 4 is dashed and is not included
in the graph of the solution. The solution region is shaded in the graph below.
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.
ANSWER: eSolutions Manual - Powered by Cognero
Page 22
The solution region is shaded in the graph below.
The solution of the system is the set of ordered pairs
in Guide
the intersection
of the- Chapter
graphs of 6y ≤ 5 and y > x −
Study
and Review
4. Overlay the graphs and locate the green region.
This is the intersection.
The graph of y ≥ −2x + 4 is solid and is included in
the graph of the solution. The solution region is shaded in the graph below.
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.
ANSWER: The solution region is shaded in the graph below.
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. ANSWER: 54. y ≤ −x − 3
y ≥ 3x − 2
The Manual
graph -ofPowered
y ≥ −2x
4 is solid and is included in
eSolutions
by +
Cognero
the graph of the solution. SOLUTION: Page 23
Graph each inequality.
The graph of y ≤ −x − 3 is solid and is included in
Study
54. y ≤Guide
−x − 3and Review - Chapter 6
y ≥ 3x − 2
ANSWER: 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.
ANSWER: eSolutions Manual - Powered by Cognero
Page 24