Upload
others
View
18
Download
0
Embed Size (px)
Citation preview
ANSWERS FOR EXERCISES
MATH GRADE 8 UNIT 5
LINEAR EQUATIONS
Copyright © 2014 Pearson Education, Inc. 22
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. x = 4
x + 3 = 2x – 1 x + 3 – x + 1 = 2x – 1 – x + 1 4 = x
2. n = –5
5n + 4 = 3(n – 2) 5n + 4 = 3n – 6 5n + 4 – 4 – 3n = 3n – 6 – 4 – 3n 2n = –10 22
–102
=n
n = –5
3. n = 15
2n + 3n – 8 = 4n + 7 5n – 8 = 4n + 7 5n – 8 + 8 – 4n = 4n + 7 + 8 – 4n n = 15
4. n = 612
or n = 6.5
3(n – 4) = 5(n – 5) 3n – 12 = 5n – 25 3n – 12 – 3n + 25 = 5n – 25 – 3n + 25 13 = 2n 132
22
= n
= n612
LESSON 2: ONE VARIABLE ON BOTH SIDES
Copyright © 2014 Pearson Education, Inc. 23
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 2: ONE VARIABLE ON BOTH SIDES
5. n = –1
2n + 6 + n = 4n – 3n + 4 3n + 6 = n + 4 3n + 6 – n – 6 = n + 4 – n – 6 2n = –2 22
–22
=n
n = –1
6. n = –125
or –1.4
6n = (n + 3) – 10 6n = n – 7 6n – n = n – 7 – n 5n = –7 55
–75
=n
n = –125
Challenge Problem
7. Answers will vary. Possible answer:
Erin and Pedra are biking the same trail and in the same direction. Erin is 3 miles from the start of the trail and bikes 3 miles per hour. Pedra is 8.6 miles from the start of the trail and bikes 2 miles per hour. When will the two girls be the same distance from the start of the trail?
3h + 3 = 2h + 8.6 3h + 3 – 2h – 3 = 2h + 8.6 – 2h – 3 h = 5.6
The two girls will be the same distance after 5.6 hr, or 5 hr 36 min.
Copyright © 2014 Pearson Education, Inc. 24
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. x = 3
x + 5 = 4(x – 1) x + 5 = 4x – 4 x + 5 – x + 4 = 4x – 4 – x + 4 9 = 3x 93
33
= x
3 = x
2. No solution
3x + 4 = 6 + 3x 3x + 4 – 3x – 4 = 6 + 3x – 3x – 4 0 ≠ 2
3. All numbers as solutions
x + 9 + 3x = 4x + 9 4x + 9 = 4x + 9 4x + 9 – 9 = 4x + 9 – 9 4x = 4x x = x
4. One solution
3x + 4(x – 5) = 5x + 8 3x + 4x – 20 = 5x + 8 7x – 20 = 5x + 8 7x – 20 – 5x + 20 = 5x + 8 – 5x + 20 2x = 28 x = 14
5. No solution
3(x + 2) = 3x + 2 3x + 6 – 3x – 6 = 3x + 2 – 3x – 6 0 ≠ –4
LESSON 3: HOW MANY SOLUTIONS?
Copyright © 2014 Pearson Education, Inc. 25
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 3: HOW MANY SOLUTIONS?
6. All numbers as solutions
2(x – 6) = –10 + 2x – 2 2x – 12 = –12 + 2x 2x – 12 + 12 = –12 + 2x + 12 2x = 2x x = x
7. One solution
5x + x + 4(x – 1) = 3x – 4 + x 5x + x + 4x – 4 = 4x – 4 10x – 4 – 4x + 4 = 4x – 4 – 4x + 4 6x = 0 x = 0
Challenge Problem
8. Answers will vary. Possible answers:
a. x + 4 = x + 5b. 2x = 3x
Copyright © 2014 Pearson Education, Inc. 26
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. 8x + 26 or 26 + 8x
4(x + 5) + 4x + 8 – 2 4x + 20 + 4x + 8 – 2 8x + 26
2. No solution
12345678
–5–6–7
–4–3–2
65432–1 1–2–3–4–5 x
y
g(x) = 4x – 6
f(x) = 4x + 6
3. No solution
123456789
10111213
–265432–1 1–2–3–4–5 x
y
f(x) = 3(x + 4)
g(x) = 3x + 7
LESSON 4: PARALLEL OR INTERSECTING?
Copyright © 2014 Pearson Education, Inc. 27
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 4: PARALLEL OR INTERSECTING?
4. All numbers as solutions
123456789
1011
65432–1–2–3–4
1–2–3–4–5 x
y
f(x) = 4x + 9
g(x) = 2 + 2(x + 1) + 2x + 5
5. One solution
123456789
10111213
–265432–1 1–2–3–4–5 x
y
f(x) = 8(x + 1)
g(x) = 3x + 8
Copyright © 2014 Pearson Education, Inc. 28
Grade 8 Unit 5: Linear Equations
EXERCISESLESSON 4: PARALLEL OR INTERSECTING?
6. One solution
12345678
65432–1–2–3–4–5–6–7
1–2–3–4–5 x
y
f(x) = 3x – 4x – 1
g(x) = 7x – 6
7. One solution
123456789
–5–6
–4–3–2
65432–1 1–2–3–4–5 x
y
g(x) = 3x + 1
f(x) = x + 612
Copyright © 2014 Pearson Education, Inc. 29
Grade 8 Unit 5: Linear Equations
EXERCISESLESSON 4: PARALLEL OR INTERSECTING?
Challenge Problem
8. The graphs will intersect at (4, 13). So, the solution to 2x + 5 = 3x + 1 is the value of x at that point, 4.
123456789
1011121314
–265432–1 1–2–3–4–5 x
y
f(x) = 2x + 5
g(x) = 3x + 1
Copyright © 2014 Pearson Education, Inc. 30
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. B 4x – y = –5
2. a. y = x – 3
b. y = –x + 5
c. y = x + 7
d. y = x + 0
3. a. 3x – y = 4
b. 12
x + y = 1
c. 2x + y = 2
4. a. Slope: –2; y-intercept: 8
2x + y = 8 written in slope-intercept form is y = –2x + 8.
b. Slope: 1; y-intercept: –4
x – y = 4 written in slope-intercept form is y = x – 4.
5. Slope-intercept form: y = –x + 4; standard form: x + y = 4
The y-intercept is (0, 4). The slope is –1.
6. 10x + 35y = 70 10x – 10x + 35y = 70 – 10x Addition property of equality 35y = –10x + 70 Combine like terms 3535
1035
7035
yx
= − + Multiplication property of equality
y = − 27
x + 2 Simplify
Challenge Problem
7. 200x + 250y = 8,000
LESSON 5: FORMS OF LINEAR EQUATIONS
Copyright © 2014 Pearson Education, Inc. 31
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. y = –2x + 4
2. (7, 17)
y = 2x + 3 y = 3x – 4 2x + 3 = 3x – 4 2x + 3 – 2x + 4 = 3x – 4 – 2x + 4 7 = x
y = 2x + 3 y = 2(7) + 3 y = 14 + 3 y = 17
3. No solution
y = 3x – 2x y – x = 4 or y = x + 4 3x – 2x = x + 4 x = x + 4 x – x = x – x + 4 0 ≠ 4
4. Infinitely many solutions
4 + y = 4x or y = 4x – 4 y = 4(x – 1) or y = 4x – 4 4x – 4 = 4x – 4
LESSON 6: A SYSTEM OF EQUATIONS
Copyright © 2014 Pearson Education, Inc. 32
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 6: A SYSTEM OF EQUATIONS
5. –56
, 512
⎞⎠⎟
⎛⎝⎜
3 = 3x + y or y = –3x + 3 – 3x + y = 8 or y = 3x + 8 –3x + 3 = 3x + 8 –3x + 3 + 3x – 8 = 3x + 8 + 3x – 8 –5 = 6x
x = – 56
y = 3x + 8
y = 3 –56
⎞⎠⎟
⎛⎝⎜
+ 8
y = –52
+ 8
y = 112
or 512
6. (9, 36)
9 = 5x – y or y = 5x – 9 –3x + y = 9 or y = 3x + 9 5x – 9 = 3x + 9 5x – 9 –3x + 9 = 3x + 9 – 3x + 9 2x = 18 x = 9
y = 3x + 9 y = 3(9) + 9 y = 27 + 9 y = 36
Challenge Problem
7. Systems will vary. Possible system:
y = 4x + 5 y = 4x – 1 4x + 5 = 4x – 1 4x + 5 – 4x – 5 = 4x – 1 – 4x – 5 0 ≠ –6
The system of the two equations has no solution.
The lines will be parallel because both lines have the same slope, 4. The line y = 4x + 5 has a y-intercept of 5. The line y = 4x – 1 has a y-intercept of – 1.
Copyright © 2014 Pearson Education, Inc. 33
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. y = 1
3(2y – 1) + 5y = 8 6y – 3 + 5y = 8 11y – 3 = 8 11y – 3 + 3 = 8 + 3 11y = 11 y = 1
2. (6, 1)
x = 4y + 2 2x – 3y = 9 2(4y + 2) – 3y = 9 8y + 4 – 3y = 9 5y = 5 y = 1 x = 4(1) + 2 y = 6
3. (2.4, 4.8) or 225
, 445
⎛⎝⎜
⎞⎠⎟
y = 2x 3x + y = 12 3x + 2x = 12 5x = 12
x = ⎛⎝⎜⎞⎠⎟2
125
or 225
y = ⎛⎝⎜
⎞⎠⎟2
125
y = 245
or 445
LESSON 7: SOLVING—SUBSTITUTION
Copyright © 2014 Pearson Education, Inc. 34
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 7: SOLVING—SUBSTITUTION
4. (–5, 20)
3x + y = 5 or y = –3x + 5 4x + 2y = 20 4x + 2(–3x + 5) = 20 4x – 6x + 10 = 20 – 2x + 10 = 20 – 2x + 10 + 2x – 20 = 20 + 2x – 20 – 10 = 2x x = –5
y = –3(–5) + 5 y = 15 + 5 y = 20
5. No solution
4x + 5 = y 4x – y = 8 4x – (4x + 5) = 8 4x – 4x – 5 = 8 – 5 ≠ 8
6. Infinitely many solutions
y = 3x – 7 + x 7 = 4x – y 7 = 4x – (3x – 7 + x) 7 = 4x – 3x + 7 – x 7 = 7
Challenge Problem
7. Answers will vary. Ask a classmate to check your system of equations.
Possible system: y = x + 1 y = –2x + 13 x + 1 = –2x + 13 x + 1 + 2x – 1 = –2x + 13 + 2x – 1 3x = 12 x = 4
y = 4 + 1 y = 5
Copyright © 2014 Pearson Education, Inc. 35
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. 5x = 1
3x – y = 7+ 2x + y = –6
5x + 0 = 1
2. (2, –1)
4x – y = 9+ 3x + y = 5
7x + 0 = 14
x = 2
3(2) + y = 5 6 + y = 5 y = –1
3. 312
, 112
⎛⎝⎜
⎞⎠⎟ or (3.5, 1.5)
x + 3y = 8+ –x + 5y = 4
0 + 8y = 12
y = 128
or 32
or 112
x + 3⎛⎝⎜
⎞⎠⎟
32
= 8
x + 92
= 8
x = 312
LESSON 8: SOLVING—ELIMINATION
Copyright © 2014 Pearson Education, Inc. 36
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 8: SOLVING—ELIMINATION
4. 3, – 213
⎛⎝⎜
⎞⎠⎟
4x + 3y = 5 or –1(4x + 3y) = –1(5) or –4x –3y = –5
5x + 3y = 8+ –4x – 3y = –5
x + 0 = 3
x = 3
5(3) + 3y = 8 15 + 3y = 8 3y = –7
y = – 73
or – 213
5. 12
,58
⎛⎝⎜
⎞⎠⎟
3x – 4y = –1 or 2(3x – 4y) = 2(–1) or 6x – 8y = –2
2x + 8y = 6+ 6x – 8y = –2
8x + 0 = 4
x = 12
2(12
) + 8y = 6
1 + 8y = 6 8y = 5 y = 5
8
Copyright © 2014 Pearson Education, Inc. 37
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 8: SOLVING—ELIMINATION
Challenge Problem
6. a. Since the first equation is solved for x, it is easy to substitute the value for x into the second equation.
b. 12023
,523
⎛⎝⎜
⎞⎠⎟
x = 4y + 1 3y + 5x = 10 3y + 5(4y + 1) = 10 3y + 20y + 5 = 10 23y = 5
y = 523
4523
1= ⎛⎝⎜
⎞⎠⎟ +x
x = 2023
+ 1
x = 12023
Copyright © 2014 Pearson Education, Inc. 38
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. The two numbers are 17 and 8.
x + y = 25 x – y = 9 2x + 0 = 34 x = 17
17 + y = 25 y = 8
2. a. Tower 1: slope = –12
; y-intercept = 5
Tower 2: slope = 1; y-intercept = –1
b. Tower 1: y = –12
x + 5
Tower 2: y = x – 1
c. The fire is at coordinates (4, 3).
123456789
10
–5–4–3–2
10987654321 x
y
Tower 1
Tower 2
Challenge Problem
3. No. The equation for Tower 3 is y = – 32
x + 712
. The point (4, 3) is not on that line.
On the line of sight for Tower 3, x = 4 and y = 112
.
LESSON 9: USING A SYSTEM OF EQUATIONS
Copyright © 2014 Pearson Education, Inc. 39
Grade 8 Unit 5: Linear Equations
ANSWERS
ANSWERS
1. y = 2x + 1
y = 2x + b 5 = 2(2) + b 5 = 4 + b 1 = b y = 2x + 1
2. a. Table A: slope = 2
4 02 0
( )( )
−− =
42
= 2
Table B: slope = 32
9 34 0
( )( )
−− =
64
= 32
b. Use slope-intercept to write the equation. (y = mx + b)
Table A: y = 2x When x = 0, y = 0, so y-intercept is 0. The slope is 2.
Table B: y = 32
x + 3
When x = 0, y = 3, so y-intercept is 3. The slope is 32
.
c. T(6, 12)
y = 2x
– y = – 32
x – 3
0 = 12
x – 3
3 = 12
x
6 = x
y = 2x y = 2(6) y = 12
LESSON 10: SOLVING PROBLEMS
Copyright © 2014 Pearson Education, Inc. 40
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 10: SOLVING PROBLEMS
3. a. Janes is less expensive for a 3-hour job.
Jones: y = $40x + $40 y = $40(3) + $40 y = $120 + $40 y = $160
Janes: y = $30x + $60 y = $30(3) + $60 y = $90 + $60 y = $150
b. 36034032030028026024022020018016014012010080604020
543 82 71 6 x
y
Hours
Dollars
Jones
Janes
c. (2, 120)
y = 40x + 40+ –y = –30x – 60
0 = 10x – 20
20 = 10x 2 = x
y = 40x + 40 y = 40(2) + 40 y = 80 + 40 y = 120
d. The cost for a 2-hour visit will be the same for either company, $120.
Copyright © 2014 Pearson Education, Inc. 41
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 10: SOLVING PROBLEMS
Challenge Problem
4. a. Segment AC: slope = 23
; Segment DB: slope = – 12
Segment AC:
3 14 2
46
23
– –– –
( )( )
= =
Segment DB:
4 –12
–112
– 512
3.5–7
–12
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
= =
b. Segment AC: y-intercept = 13
; Segment DB: y-intercept = 314
Segment AC:
–1 = 23
(–2) + b
–1 = –43
+ b
13
= b
Segment DB:
12
–12
512
= ⎛⎝⎜
⎞⎠⎟ + b
12
= – 114
+ b
134
= b
b = 314
c. y = 23
x + 13
y = –12
x + 314
(continues)
Copyright © 2014 Pearson Education, Inc. 42
Grade 8 Unit 5: Linear Equations
ANSWERSLESSON 10: SOLVING PROBLEMS
(continued)
d. ⎛⎝⎜
⎞⎠⎟2
12
, 2
y = 23
x + 13
–y = 12
x – 314
0 = 76
x – 3512
3512
= 76
x
52
= x or x = 212
y = 23
x + 13
y = ⎛⎝⎜
⎞⎠⎟
+23
52
13
y = 53
+ 13
y = 63
or 2