CHAPTER Solutions Key
2 Linear Functions
ARE YOU READY? PAGE 87 24. 1000 mL = 1 L
_
1000 mL = _50 mL
1. C 2. A 1L xL
1000x = 50
3. B 4. E
_
1000x = _ 50
5. x + 4y = 25 6. 1000 1000
3x - 20 > 10 x = 0.05 L
7. _x <_
15 25. 16 - 22 26. 32 - 20
8. -8 + x = -20
12 x = -6 = 12
+8
______ +8
____
=6 = 12
x = -12
27. 8 - 17 + 9 28. -0.75 + 0.625
9. -12 = -3x 10. x – 19 = -12
= -9 + 9 = -0.125
_
-12 = _ -3x + 19 ____
_____ + 19
=0 = 0.125
-3 -3 x= 7
4=x
11. 0.75 = _
x 12. 15 = 0.3x 2-1 SOLVING LINEAR EQUATIONS AND
5 _15 = _0.3x INEQUALITIES, PAGES 90–96
(0.75)5 = _
x(5)
0.3 0.3
5 50 = x CHECK IT OUT!
3.75 = x
1. Let c represent the number of additional cups
13. x = 0.4(140) 14. x(140) = 105 3.25 + 0.25c = 14
= 56 _
x(140) _
= 105 -3.25
____________ -3.25
_____
140 140
x = 0.75
_
0.25c = _ 10.75
0.25 0.25
x = 75% c = 43
15. x = 1.50(90) 16. 1 gallon = 4 quarts You can stack an additional 43 + 1 = 44 cups
_ between the two shelves.
=_
= 135 1 gallon x gallons
4 quarts 12 quarts 2a. 3(2 - 3p) = 42 b. -3(5 - 4r) = -9
4x = 12 _
3(2 - 3p) _
= 42 _
-3(5 - 4r) _
= -9
_
4x = _12 3 3 -3 -3
4 4 2 - 3p = 14 5 - 4r = 3
x = 3 gal -2
_______ -2
___ -5
_______ -5
___
17. 1 yard = 3 feet 18. 60 minutes = 1 hour _
-3p
=_ 12 _
-4r = _-2
_
1 yard _ x yards ______
60 min = _____
x min -3 -3 -4 -4
r=_
= p = -4 1
3 feet 15 feet 1h 1.5 h
x = 90 min 2
3x = 15
_
3x = _
15 3. 3(w + 7) - 5w = w + 12
3 3 3w + 21 - 5w = w + 12
x = 5 yd 21 - 2w = w + 12
19. 1 gallon = 4 quarts 20. 1 yard = 3 feet -w
________ -w
________
_
1 gallon
= _
3.5 gallons _
1 yard _
=
17 yards 21 - 3w = 12
4 quarts x quarts 3 feet x feet - 21
_________ - 21
____
x = 14 qt x = 51 ft ____
-3w = ___
-9
21. 60 minutes = 1 hour -3 -3
_
60 minutes = __
200 minutes w=3
1 hour x hours 4a. 5(x - 6) = 3x - 18 + 2x
60x = 200 5x - 30 = 5x - 18
_60x = _
200 -5x
________ -5x
________
60 60 -30 = -18 ✗
x = 3_
1 h or 3 h 20 min
no solution, ∅
3
b. 3(2 - 3x) = -7x - 2(x - 3)
22. 100 cm = 1 m 23. 1 km = 1000 m
6 - 9x = -7x - 2x + 6
_
100 cm = _107 cm _1 km = _ 2.5 km
6 - 9x = -9x + 6
1m xm 1000 m xm
100x = 107 x = 2500 m + 9x _______
______ + 9x
_
100x = _107 6=6✓
100 100 all real numbers, !
x = 1.07 m
35 Holt McDougal Algebra 2
, 5. x + 8 ≥ 4x + 17 9. x = -2(x - 3) 10. (t - 3)7 = 6t + 21
_______
-4x ________
-4x x = -2x + 6 7t - 21 = 6t + 21
-3x + 8 ≥ 17 + 2x _______
____ + 2x - 6t
________ - 6t
________
- 8 ___
_______ -8 _3x = _
6 t - 21 = 21
-3x ≥ 9 3 3 + 21 ____
_____ + 21
_
-3x ≤ _ 9 x=2 t = 42
-3 -3
x ≤ -3 11. -0.5(r - 2) = -r - 2 12. -4t + 1 = 3t + 1 - 7t
-0.5r + 1 = -r - 2 -4t + 1 = -4t + 1
+r
________ + r
______ + 4t
_______ + 4t
_______
THINK AND DISCUSS
0.5r + 1 = -2 1=1✓
1. Possible answers: 3x - 3x = 1; 3x = 3x = 0 -1 -
_______ ___1 !
2. Multiplying by a negative number reflects the _
0.5r = _-3
numbers across 0 on the number line, thus 0.5 0.5
r = -6
reversing their positions from left to right.
-3 < 3 is true because -3 is left of 3 on the number 13. 2(3x + 1) = 3(2x + 1) 14. 2(3n + 3) - 9 = 6n
line. Unless the inequality symbol is reversed, the 6x + 2 = 6x + 3 6n + 6 - 9 = 6n
resulting statement is false. - 6x
_______ - 6x
_______ 6n + 3 = 6n
2=3✗ - 6n
_______ - 6n
____
3. 3IMILARITIES $IFFERENCES
∅ 3=0✗
9OU ADD OR SUBTRACT 7HEN YOU MULTIPLY
THE SAME QUANTITY ON OR DIVIDE BY A ∅
BOTH SIDES NEGATIVE NUMBER
9OU MULTIPLY OR YOU REVERSE THE 15. 2h + 4 - 5h = -3h + 4
3OLVING %QUATIONS INEQUALITY SIGN
DIVIDE THE SAME
POSITIVE NUMBER
AND )NEQUALITIES
)NEQUALITIES USUALLY 4 - 3h = -3h + 4
ON BOTH SIDES HAVE AN INFINITE ______
+ 3h _______
+ 3h
NUMBER OF SOLUTIONS
WHILE EQUATIONS 4=4✓
USUALLY HAVE
!
16. 4(2 - 6m) = 6(2 - 4m)
EXERCISES 8 - 24m = 12 - 24m
+ 24m
_______ + 24m
________
GUIDED PRACTICE
8 = 12 ✗
1. identity ∅
2. Let d represent the number of DVDs she can rent. 17. 0.5(-8p + 1) = -4p + 1
4.95 + 1.95d = 42 -4p + 0.5 = -4p + 1
- 4.95
____________ - 4.95
_____ + 4p
_________ + 4p
_______
_
1.95d = _ 37.05 0.5 = 1 ✗
1.95 1.95 ∅
d = 19
Shanti can rent 19 DVDs in her first month. 18. 5x – 12 > 8
+ 12 ____
______ + 12
3. 8(x - 5) = 72 4. 1.5(x - 4) = 9.6
5x > 20
8x - 40 = 72 1.5x - 6 = 9.6
+ 40 ____
______ + 40 + 6 ____
______ +6
_5x > _
20
5 5
_
8x = _
112 _
1.5x = _15.6 x>4
8 8 1.5 1.5
x = 14 x = 10.4
5. -27 = 3(x - 3) 6. 5 - 4c = c + 20
19. 62 - 18x < 20
-27 = 3x - 9 + 4c ________
______ + 4c
- 62
_________ - 62
____
+9
____ +9
_____ 5 = 5c + 20
-18x < -42
_
-18 = _
3x - 20
____ - 20
_______
_
-18x > _
-42
3 3 _
-15 = _5c
-18 -18
-6 = x
x>_
5 5 7
-3 = c 3
7. 24 + 7x = -4x - 9 8. 3(x - 5) = 5x + 9
+ 4x _______
_______ + 4x 3x - 15 = 5x + 9
24 + 11x = -9 ________
-3x - 3x
_______
- 24
_________ - 24
____ -15 = 2x + 9
_
11x = _-33 -9
____ -9
______
11 11 _-24 = _2x
x = -3 2 2
-12 = x
, 36. 4x - 2(3 + 2x) = -6 42. Let x represent the number of touchdown passes
4x - 6 - 4x = -6 Brandon Stokely caught.
-6 = -6 ✓ x + (x + 2) + ((x + 2) + 3) = 37
! x + x + 2 + x + 2 + 3 = 37
37. -3x + 8 ≤ 14 3x + 7 = 37
- 8 ___
_______ -8 - 7 ___
______ -7
-3x ≤ 6 _
3x = _30
_
-3x ≥ _ 6 3 3
x = 10
-3 -3
x ≥ -2 Brandon Stokely caught 10, Reggie Wayne caught 12,
and Marvin Harrison caught 15 touchdown passes.
43a. Let m represent the number of 3-minute messages.
3m + 0.5 ≤ 32
38. 3(x - 1)> 7(x + 3)
- 0.5 ____
________ - 0.5
3x - 3 > 7x + 21
- 7x
_______ - 7x
________
_3m ≤ _31.5
3 3
-4x - 3 > 21 m ≤ 10.5
+ 3 ___
_______ +3 The machine can record no more than ten 3-minute
-4x > 24 messages.
_
-4x < _24
b. Let m represent the number of messages.
-4 -4
x < -6 1.5m + 0.5 ≤ 32
- 0.5 _____
_________ - 0.5
_____
1.5m ≤ ____
31.5
39. 5(x - 2) ≥ 4(2x + 6) + 2 1.5 1.5
5x - 10 ≥ 8x + 24 + 2 m ≤ 21
5x - 10 ≥ 8x + 26 The machine can record no more than 21 average-
- 8x
________ - 8x
_________ length messages.
-3x - 10 ≥ 26 c. Let m represent the number of messages.
+ 10 ____
_______ + 10 3(2) + 5 + 1.5m + 0.5 = 32
-3x ≥ 36 6 + 5 + 1.5m + 0.5 = 32
_
-3x ≤ _ 36 11.5 + 1.5m = 32
-3 -3 - 11.5
____________ - 11.5
_____
x ≤ -12
_
1.5m = _
20.5
1.5 1.5
m = 13_2
40. Let v represent the value of the sales. 3
The machine can record 13 more average-length
500 + 0.15v ≥ 2000
messages.
- 500
___________ - 500
_____
0.15v ≥ 1500 44. x + 2x + (x – 20) = 180
_
0.15v ≥ _
1500 4x – 20 = 180
0.15 0.15 + 20 ____
______ + 20
v ≥ 10,000 _4x = _
200
Pat must sell $10,000 worth of jewelry in a month to 4 4
earn at least $2000. x = 50
m∠A = 50°, m∠B = 100°, m∠C = 30°
41. Let x represent the cost of a loaf of bread in 1902.
44x = 1.72 + x 45. 90 + 3.5x + x = 180
- x _______
___ -x 90 + 4.5x = 180
_=_
43x 1.72 - 90
__________ - 90
____
43 43 _
4.5x = _
90
x = 0.04 4.5 4.5
A loaf of bread cost $0.04 in 1902. x = 20
m∠D = 70°, m∠E = 20°, m∠F = 90°
Possible answer:
Estimate the cost of 44 loaves in 1902:
44($0.04) ≈ 40($0.05) = $2.00.
Find the cost of 1 loaf in 2006:
$1.72 + $0.04 = $1.76.
The estimated cost of 44 loaves in 1902 ≈ the cost
of 1 loaf in 2006.
38 Holt McDougal Algebra 2
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