unit 7 review for final
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Which linear system has the solution x = –2 and y = 6?
a. x + 3y = 16
c. x + 2y = –2
4x + 4y = 16
2x + 4y = –4
b. x + 3y = 17
d. 2x + y = –2
2x + y = 15
x + y = 16
____
2. Which linear system has the solution x = 4 and y = –2?
a. x + 4y = 15
c. 4x + y = 14
4x
= –17
–2x
= –16
b. 2x + 4y = 4
d. x + 4y = 4
–2x + y = 14
2x + 4y = 8
____
3. Which linear system has the solution x = 8 and y = 2.5?
a. 2x + 2y = 21
c. 2x + 2y = 8
2x – 2y = 11
x – y = 21
b. x + 2y = 8
d. x + 3y = 22
2x – 4y = 16
2x – y = 10
____
4. Create a linear system to model this situation:
The perimeter of an isosceles triangle is 36 cm. The base of the triangle is 9 cm longer than each equal side.
a. s + b = 36
b. 2s + b = 36
c. 2b + s = 36
d. 2s + b = 36
b–9=s
b+9=s
s+9=b
s+9=b
____
5. Create a linear system to model this situation:
A collection of nickels and dimes contains four times as many dimes as nickels. The total value of the
collection is $20.25.
a. d = 4n
b. d = 4n
c. n = 4d
d. d + n = 15
5n + 10d = 2025
5d + 10n = 2025
5n + 10d = 2025
5n + 10d = 2025
____
6. Create a linear system to model this situation:
In a board game, Judy scored 3 points more than twice the number of points Ann scored.
There was a total of 39 points scored.
a. j = 3 + 2a
b. j – 3 = 2a
c. j + 3 = 2a
d. a = 3 + 2j
j + a = 39
j + 2a = 39
j + a = 39
j + a = 39
____
7. Create a linear system to model this situation:
A woman is 3 times as old as her son. In thirteen years, she will be 2 times as old as her son will be.
a. w = s + 3
c. w = 3s
w + 13 = 2s
w = 2s
b. w = 3s
d. w = 3s
w + 13 = 2(s + 13)
s + 13 = 2(w + 13)
____
8. Create a linear system to model this situation:
Cheri operates a grass-cutting business. She charges $19 for a small lawn and $29 for a large lawn. One
weekend, Cheri made $287 by cutting 13 lawns.
a. s + l = 13
c. s + l = 13
19s + 29l = 287
29s + 19l = 287
b. s + l = 287
19s + 29l = 13
____
d. s + l = 287
29s + 19l = 13
9. Create a linear system to model this situation:
A length of outdoor lights is formed from strings that are 5 ft. long and 11 ft. long. Fourteen strings of lights
are 106 ft. long.
a. 5x + 11y = 14
c. x + y = 14
x + y = 106
5x + 11y = 106(14)
b. x + y = 14
d. x + y = 14
5x + 11y = 106
x + 2y = 106
____ 10. Create a linear system to model this situation:
A rectangular field is 35 m longer than it is wide. The length of the fence around
the perimeter of the field is 290 m.
a. l + 35 = w
b. l = w + 35
c. l = w + 35
d. l = w + 35
2l + 2w = 290
2l + 2w = 290
l + w = 290
lw = 290
____ 11. Create a linear system to model this situation:
Tickets for a school play cost $8 for adults and $4.75 for students.
There were ten more student tickets sold than adult tickets, and a total of $1399 in ticket sales was collected.
a. 8a + 4.75s = 1399
c. 8a + 4.75s = 1399
s = a + 10
a = s + 10
b. 8a + 4.75s = 1399
d. 4.75a + 8s = 1399
a + s = 10
s = a + 10
____ 12. Match each situation to a linear system below.
A. The perimeter of a rectangular playground is 163 m. The length is 6 m less than double
the width.
B. The perimeter of a rectangular playground is 163 m. The width is one-half the length
decreased by 6 m.
C. The perimeter of a rectangular playground is 163 m. The length decreased by 6 m is
double the width.
i)
a. A-i, B-ii, C-iii
b. A-iii, B-i, C-ii
ii)
iii)
c. A-ii, B-i, C-iii
d. A-i, B-iii, C-ii
____ 13. Which graph represents the solution of the linear system:
y = –2x
y + 6 = 2x
Graph A
–6
–4
–2
6
4
4
2
2
0
2
–2
–4
4
6
x
–4
–2
0
–4
–6
–6
6
4
4
2
2
2
–2
4
6
x
–6
–4
–2
(1, –2)
6
x
6
x
6
x
(2.2, –0.5)
0
2
–2
–4
–4
–6
–6
a. Graph B
b. Graph A
4
y
Graph D
6
0
2
–2
(2, –2)
y
–2
–6
–4
Graph B
–6
y
Graph C
y
6
4
(1.4, –0.8)
c. Graph C
d. Graph D
____ 14. Which graph represents the solution of the linear system:
–3x – y = –5
4x – y =
y
Graph A
–6
–4
–2
y
Graph C
6
6
4
4
2
(1, 2)
0
2
4
6
x
–6
–4
–2
2
(–1, –2)
0
2
–2
–2
–4
–4
–6
–6
4
y
Graph B
6
4
4
2
–6
–4
–2
y
Graph D
6
2
(0, 0)
0
2
4
6
x
–6
–4
–2
0
–2
–2
–4
–4
–6
–6
a. Graph A
b. Graph B
c. Graph C
d. Graph D
____ 15. Use the graph to solve the linear system:
y = –3x – 5
y
= 3x
y
6
4
y –1 =3 x
2
–6
–4
–2
0
2
4
6
x
–2
–4
y = –3x – 5
–6
a. (1, –2)
b. (–1, 0)
____ 16. Use the graph to solve the linear system:
y = –5x
y + = 2x
c. (1, 0)
d. (–1, –2)
2
4
(–0.8, –2.7)
6
x
y
6
4
y = –5x –2
2
–6
–4
–2
y +2 = 2 x
0
2
4
6
x
–2
–4
–6
a. (2, 0)
b. (2, –2)
c. (0, 0)
d. (0, –2)
____ 17. Use the graph to approximate the solution of the linear system:
y
4
2
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
a. (–3, 0.2)
b. (0, –2.8)
c. (0.2, –3)
d. (–2.8, 0)
____ 18. Car A left Calgary at 8 A.M. to travel 500 mi. to Regina, at an average speed of 63 mph.
Car B left Regina at the same time to travel to Calgary at an average speed of
37 mph. A linear system that models this situation is:
d = 500 – 63t
d = 37t,
where d is the distance in miles from Regina, and t is the time in hours since 8 A.M. Which graph would you
use to determine how far the cars are from Regina when they meet? What is this distance?
Graph B
500
Distance from Regina (mi.)
Distance from Regina (mi.)
Graph A
Car A
400
Car B
300
200
(5, 185)
400
300
Car A
200
Car B
(2.5,92.5)
100
100
0
500
2
4
6
8
10
12
0
14
2
4
8
Graph D
500
Distance from Regina (mi.)
Distance from Regina (mi.)
Graph C
Car A
400
Car B
300
(4.1,195.8)
200
100
0
6
Tim e (h)
Tim e (h)
Car B
500
Car A
400
300
(3.8,200)
200
100
2
4
6
8
10
12
0
Tim e (h)
a. Graph C:
195.8 mi.
b. Graph D:
200 mi.
2
4
6
Tim e (h)
c. Graph A:
185 mi.
____ 19. Which linear system is represented by this graph?
d. Graph B:
92.5 mi.
8
10
10
a) x – y = 3
6x + 5y = 14
b)
y
8
x+y=5
6x + 5y = 14
6
4
c) x + y = 7
7x + 5y = 14
2
d) x + y = 9
5x + 6y = 14
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
–6
–8
a. System a
b. System b
c. System c
d. System d
____ 20. Which linear system is represented by this graph?
a) x – y = 5
5x + 6y = 18
y
8
b)
x–y=7
5x + 6y = 18
6
4
c) x – y = 9
6x + 6y = 18
d) x – y = 11
6x + 5y = 18
2
–8
–6
–4
–2
0
–2
–4
–6
–8
2
4
6
8
x
a. System d
b. System b
c. System a
d. System c
____ 21. Which linear system is represented by this graph?
a) 2x – 5y = –16
x=1
y
8
b)
c)
2x + 5y = 16
2x – 5y = 16
6
4
2x – 5y = 16
2
x – y = –1
5
2
d) 2x + 5y = 16
x = –1
–8
–6
–4
0
–2
2
4
6
8
–2
–4
–6
–8
a. System a
b. System d
c. System b
____ 22. Determine the solution of the linear system represented by this graph.
d. System c
x
a) (2, 3.8)
y
b) (3.8, 2)
8
c) (–3, 3.8)
6
d) (–2, 3.8)
4
2
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
–6
–8
a. b
b. a
c. d
d. c
____ 23. Two life insurance companies determine their premiums using different formulas:
Company A: p = 2a + 24
Company B: p = 2.25a + 13, where p represents the annual premium, and a represents the client’s age.
Use the graph to determine the age at which both companies charge the same premium.
p
250
Premium ($)
200
150
Company A
100
Company B
50
a
0
10
20
30
40
50
Age (years)
60
70
80
90
a. 62 years
b.
24 years
c. 59 years
d. 44 years
____ 24. At a skating rink, admission is $4.00 for a student and $8.00 for an adult.
Tuesday evening, 20 people used the skating rink and a total of $132 in admission fees was collected. A linear
system that models this situation is:
4s + 8a = 132
s + a = 20
where s represents the number of student admissions, and a represents the number of adult admissions
purchased.
Use the graph to solve this problem:
How many students used the skating rink on Tuesday evening?
Number of adult admissions
a
50
40
30
20
10
s
0
10
20
30
40
50
Num ber of student adm issions
a. 19 students
b. 20 students
c. 13 students
____ 25. Use the graph to approximate the solution of this linear system:
6x – 7y = –4
3
– y = 3x + 7
5
d. 7 students
y
8
6
4
2
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
–6
–8
a. (–0.1, 3.8)
b. (–2.1, –1.2)
c. (–1.2, 3.8)
d. (–2.1, –0.1)
c. (–6, –20)
d. (–6, 20)
c. (4, 30)
d. (–4, –30)
____ 26. Use substitution to solve this linear system.
y=
–
x
13x + 5y = 178
a. (6, –20)
b. (6, 20)
____ 27. Use substitution to solve this linear system.
x = 2y – 56
5x + 13y = 410
a. (4, –30)
b. (–4, 30)
____ 28. Identify two like terms and state how they are related.
–10x + 20y = 460
30x + 60y = 1620
a. –10x and 30x; by a factor of –3
b. –10x and 20y; by a factor of –2
c. 30x and 60y; by a factor of 2
d. –10x and 460; by a factor of 46
____ 29. Identify two like terms and state how they are related.
a.
5
7
b. 8x and –96; by a factor of 12
7x and –5y; by a factor of
c.
d.
1
2
8x and –4y; by a factor of
8x and 7x; by a factor of
7
8
____ 30. Use substitution to solve this linear system.
x=4+y
4x + 16y = –264
a. (–14, –14)
b. (–10, –10)
c. (–10, –14)
d. (–14, –10)
____ 31. Use substitution to solve this problem:
The perimeter of a rectangular field is 276 m. The length is 18 m longer than the width.
What are the dimensions of the field?
a. 58 m by 80 m
b. 68 m by 70 m
c. 78 m by 60 m
d. 48 m by 90 m
____ 32. For each equation, identify a number you could multiply each term by to ensure that the coefficients of the
variables and the constant term are integers.
5
1
47
(1)
x+ y=
4
6
12
4
6
(2)
x – y = 16
5
7
a.
b.
c.
d.
Multiply equation (1) by 35; multiply equation (2) by 12.
Multiply equation (1) by 12; multiply equation (2) by 35.
Multiply equation (1) by 2; multiply equation (2) by 3.
Multiply equation (1) by 3; multiply equation (2) by 2.
____ 33. Write an equivalent system with integer coefficients.
3
438
x + 3y =
7
7
5
310
x + 5y =
6
3
a. 3x + 21y =
5x + 30y =
b. 21x + 3y =
5x + 30y =
438
620
438
620
c. 3x + 21y = 438
30x + 5y = 620
d. 3x + 21y = 1
5x + 30y = 1
____ 34. Write an equivalent system with integer coefficients.
3
5x + y = 14
2
5
755
x + 5y =
6
6
a. 10x + 3y = 1
5x + 30y = 1
b. 3x + 10y = 28
5x + 30y = 755
c. 10x + 3y =
30x + 5y =
d. 10x + 3y =
5x + 30y =
28
755
28
755
____ 35. The solution of this linear system is (–3, y). Determine the value of y.
x – 3y = 33
6
88
x–y=
7
7
a. 20
b. 30
c. 10
d. 40
____ 36. Use an elimination strategy to solve this linear system.
a.
b.
c.
d.
and
and
and
and
____ 37. Use an elimination strategy to solve this linear system.
a.
b.
c.
d.
and
and
and
and
____ 38. Write an equivalent linear system where both equations have the same x-coefficients.
a.
b.
and
and
c.
d.
and
and
____ 39. Write an equivalent linear system where both equations have the same y-coefficients.
a.
b.
and
and
c.
d.
and
and
____ 40. Model this situation with a linear system:
Frieda has a 13% silver alloy and a 31% silver alloy. Frieda wants to make 26 kg of an alloy that is 47%
silver.
a.
and
c.
and
b.
and
d.
and
____ 41. Use an elimination strategy to solve this linear system.
a.
b.
and
and
c.
d.
____ 42. Use an elimination strategy to solve this linear system.
and
and
a.
b.
c.
d.
and
and
and
and
____ 43. Without graphing, determine which of these equations represent parallel lines.
i) –6x + 6y = 12
ii) –4x + 6y = 12
iii) –2x + 6y = 12
iv) –6x + 6y = 14
a. ii and iii
b. i and ii
c. i and iv
d. i and iii
____ 44. Determine the number of solutions of the linear system:
2x – 5y = 23
–6x + 15y = 21
a. one solution
b. no solution
c. two solutions
d. infinite solutions
____ 45. Determine the number of solutions of the linear system:
14x – 5y = 123
14x – 5y = 73
a. no solution
b. infinite solutions
c. two solutions
d. one solution
____ 46. Determine the number of solutions of the linear system:
14x + 7y = 315
16x – 2y = 610
a. no solution
b. one solution
c. two solutions
d. infinite solutions
____ 47. Determine the number of solutions of the linear system:
5x + 7y = 76
–25x – 35y = –380
a. 2 solutions
b. one solution
c. infinite solutions
d. no solution
____ 48. The first equation of a linear system is 2x + 3y = 52. Choose a second equation to form a linear system with
infinite solutions.
i) 2x + 3y = –260
ii) –10x – 15y = –260 iii) –10x + 3y = –260 iv) –10x + 3y = 255
a. Equation iii
b. Equation iv
c. Equation i
d. Equation ii
____ 49. The first equation of a linear system is 8x + 13y = 166. Choose a second equation to form a linear system with
exactly one solution.
i) 8x + 13y = –830
ii) –40x – 65y = –830 iii) –40x + 13y = –830 iv) –40x – 65y = 0
a. Equation iii
b. Equation i
c. Equation ii
d. Equation iv
____ 50. The first equation of a linear system is –6x + 12y = –42. Choose a second equation to form a linear system
with no solution.
i) –6x + 12y = 126
ii) 18x – 36y = 126
iii) 18x + 12y = 126
iv) 18x + 36y = 0
a. Equation iv
b. Equation ii
c. Equation iii
d. Equation i
____ 51. Two lines in a linear system have the same slope, but different y-intercepts.
How many solutions does the linear system have?
a. two solutions
b. no solution
c. infinite solutions
d. one solution
Short Answer
52. Quincy used this linear system to represent a situation involving a collection of $5 bills and $10 bills:
f + t = 70
5f + 10t = 575
a) What problem might Quincy have written?
b) What does each variable represent?
53. Solve this linear system by graphing.
–3x – 2y = 16
–x + y = –8
y
8
6
4
2
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
–6
–8
54. a)
Write a linear system to model this situation:
A hockey coach bought 25 pucks for a total cost of $70. The pucks used for practice cost
$2.50 each, and the pucks used for games cost $3.25 each.
b) Use a graph to solve this problem:
How many of each type of puck did the coach purchase?
g
Pucks used for games
50
40
30
20
10
p
0
10
20
30
40
50
Pucks used for practice
55. Use substitution to solve this linear system:
56. Use substitution to solve this linear system:
7
x + y = –34
8
–3x + 4y = –4
57. Create a linear system to model this situation. Then use substitution to solve the linear system to solve the
problem.
At the local fair, the admission fee is $8.00 for an adult and $4.50 for a youth. One Saturday, 209 admissions
were purchased, with total receipts of $1304.50. How many adult admissions and how many youth
admissions were purchased?
58. Determine the number of solutions of this linear system.
7x – 3y = 43
7x – 3y = 13
59. Determine the number of solutions of this linear system.
15x + 30y = –240
17x + 21y = 53
Problem
60. a) Write a linear system to model this situation:
The coin box of a vending machine contains $23.75 in quarters and loonies. There are 35 coins in all.
b) Use a graph to solve this problem:
How many of each coin are there in the coin box?
61. a) Write a linear system to model this situation.
Mrs. Cheechoo paid $155 for one-day tickets to Silverwood Theme Park for herself, her husband, and 3
children. Next month, she paid $285 for herself, 3 adults, and 5 children.
b) Use a graph to solve this problem:
What are the prices of a one-day ticket for an adult and for a child?
62. a) Write a linear system to model the situation:
For the school play, the cost of one adult ticket is $6 and the cost of one student ticket is $4. Twice as
many student tickets as adult tickets were sold. The total receipts were $2016.
b) Use substitution to solve the related problem:
How many of each type of ticket were sold?
63. Use an elimination strategy to solve this linear system. Verify the solution.
64. a) Model this situation with a linear system:
To rent a car, a person is charged a daily rate and a fee for each kilometre driven. When Chena rented a
car for 15 days and drove 800 km, the charge was $715.00. When she rented the same car for 25 days and
drove 2250 km, the charge was $1512.50.
b) Determine the daily rate and the fee for each kilometre driven. Verify the solution.
65. Use an elimination strategy to solve this linear system. Verify the solution.
66. Explain what happens when you try to solve this linear system using an elimination strategy. What does this
tell you about the graphs of these equations?
67. Explain what happens when you try to solve this linear system using a substitution strategy. What does this
indicate about the graphs of these equations?
unit 7 review for final
Answer Section
MULTIPLE CHOICE
1. ANS:
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2. ANS:
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14. ANS:
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15. ANS:
A
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Moderate
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Moderate
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Moderate
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Moderate
7.1 Developing Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
16.
17.
18.
19.
20.
21.
22.
23.
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7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC:
32. ANS:
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46. ANS:
REF:
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47. ANS:
REF:
TOP:
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Easy
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
B
PTS: 1
DIF: Easy
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
D
PTS: 1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
A
PTS: 1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
KEY: Procedural Knowledge
C
PTS: 1
DIF: Easy
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Easy
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Easy
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
C
PTS: 1
DIF: Easy
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
48. ANS:
REF:
TOP:
49. ANS:
REF:
TOP:
50. ANS:
REF:
TOP:
51. ANS:
REF:
TOP:
D
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
A
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
D
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
B
PTS: 1
DIF: Moderate
7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
Relations and Functions
KEY: Conceptual Understanding
SHORT ANSWER
52. ANS:
a) There are 70 bills in a collection of $5 bills and $10 bills.
The value of the collection of bills is $575.
How many $5 bills and $10 bills are in the collection?
b) Variable f represents the number of $5 bills, and variable t represents the number of $10 bills.
PTS: 1
LOC: 10.RF9
53. ANS:
(0, –8)
DIF: Moderate
REF: 7.1 Developing Systems of Linear Equations
TOP: Relations and Functions
KEY: Conceptual Understanding
y
8
6
4
2
–8
–6
–4
–2
0
2
4
6
8
x
–2
–4
–6
–8
PTS: 1
LOC: 10.RF9
54. ANS:
a)
DIF: Easy
REF: 7.2 Solving a System of Linear Equations Graphically
TOP: Relations and Functions
KEY: Conceptual Understanding
b)
p + g = 25
2.5p + 3.25g = 70
The team purchased 15 pucks for practice and
10 pucks for games.
Pucks purchased for games
g
50
40
30
20
10
p
0
10
20
30
40
50
Pucks purchased for practice
PTS: 1
LOC: 10.RF9
55. ANS:
x = –55; y = –18
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
TOP: Relations and Functions
KEY: Conceptual Understanding
PTS: 1
DIF: Moderate
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
56. ANS:
x = –20; y = –16
PTS: 1
DIF: Moderate
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
57. ANS:
Let a represent the number of adult admissions, and y represent the number of youth admissions purchased.
a + y = 209
8a + 4.5y = 1304.5
104 adult admissions and 105 youth admissions were purchased.
PTS: 1
DIF: Moderate
REF: 7.4 Using a Substitution Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Conceptual Understanding
58. ANS:
No solutions
PTS: 1
LOC: 10.RF9
DIF: Easy
REF: 7.6 Properties of Systems of Linear Equations
TOP: Relations and Functions
KEY: Conceptual Understanding
59. ANS:
One solution
PTS: 1
LOC: 10.RF9
DIF: Easy
REF: 7.6 Properties of Systems of Linear Equations
TOP: Relations and Functions
KEY: Conceptual Understanding
PROBLEM
60. ANS:
a) Let q represent the number of quarters, and l represent the number of loonies.
The value of q quarters is 25q cents, and the value of l loonies is 100l cents.
Then, a system of equations is:
q + l = 35
25q + 100l = 2375
b)
l
Number of loonies
50
40
30
20
10
q
0
10
20
30
40
50
Num ber of quarters
Since the intersection point is at (15, 20), there are 15 quarters and 20 loonies in the coin box.
PTS: 1
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
LOC: 10.RF9
TOP: Relations and Functions
KEY: Problem-Solving Skills
61. ANS:
a) Let a represent the cost in dollars for a one-day adult ticket, and c represent the cost in dollars for a
one-day child ticket.
Then, a system of equations is:
2a + 3c = 155
4a + 5c = 285
b)
c
Cost of child's ticket ($)
50
40
30
20
10
a
0
10
20
30
40
50
Cost of adult's ticket ($)
Since the intersection point is at (40, 25), the cost of a one-day adult ticket is $40, and the cost of a
one-day child ticket is $25.
PTS: 1
LOC: 10.RF9
62. ANS:
DIF: Moderate
REF: 7.2 Solving a System of Linear Equations Graphically
TOP: Relations and Functions
KEY: Problem-Solving Skills
a) Let a represent the number of adult tickets sold, and s represent the number of student tickets sold.
There were twice as many student tickets as adult tickets.
The first equation is:
2a = s
The total receipts were $2016.
The second equation is:
6a + 4s = 2016
The linear system is:
2a = s
(1)
6a + 4s = 2016
(2)
b) Solve for s in equation (1).
2a = s
(1)
s = 2a
Substitute s = 2a in equation (2).
6a + 4s = 2016 (2)
6a + 4(2a) = 2016
6a + 8a = 2016
14a = 2016
a=
a = 144
Substitute a = 144 in equation (1).
2a = s
(1)
2(144) = s
288 = s
144 adult tickets and 288 student tickets were sold.
PTS:
REF:
LOC:
KEY:
63. ANS:
1
DIF: Moderate
7.4 Using a Substitution Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
Problem-Solving Skills | Communication
Multiply equation ‚ by 2, then subtract to eliminate x.
2 equation :
Subtract equation ƒ from equation
Substitute
in equation
.
.
Verify the solution.
In each equation, substitute:
and
For each equation, the left side is equal to the right side, so the solution is:
PTS:
REF:
LOC:
KEY:
1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
Communication | Problem-Solving Skills
and
64. ANS:
a) Let d dollars represent the daily rate and let k dollars represent the fee for each kilometre driven.
The linear system is:
b) Multiply equation
25 equation :
15
equation
by 15, then subtract to eliminate d.
:
Subtract equation
Substitute
by 25 and equation
from equation
in equation
Verify the solution.
In each equation, substitute:
.
.
and
So, the daily rate is $29 and the fee for each kilometre driven is $0.35.
PTS: 1
DIF: Difficult
REF: 7.5 Using an Elimination Strategy to Solve a System of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Problem-Solving Skills
65. ANS:
Multiply equation
7 equation :
by 7, then add to eliminate y.
Add:
Substitute
in equation
Verify the solution.
In each equation, substitute:
.
and
For each equation, the left side is equal to the right side, so the solution is:
PTS:
REF:
LOC:
KEY:
66. ANS:
1
DIF: Moderate
7.5 Using an Elimination Strategy to Solve a System of Linear Equations
10.RF9
TOP: Relations and Functions
Communication | Problem-Solving Skills
Eliminate x first.
Multiply equation
3 equation :
by 3, then add.
and
When I try to eliminate one variable, I eliminate the other variable and the constant term, so the equations
must be equivalent. This indicates that the graphs of these equations are coincident lines. So, the linear system
has infinite solutions.
PTS: 1
DIF: Moderate
REF: 7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills
67. ANS:
Solve equation
for y:
Substitute
does not equal
in equation
.
, so the linear system has no solution. This tells me that the graphs of these equations
are parallel.
PTS: 1
DIF: Difficult
REF: 7.6 Properties of Systems of Linear Equations
LOC: 10.RF9
TOP: Relations and Functions
KEY: Communication | Problem-Solving Skills