# Algebra Rules Review Sheet Page 23

REVIEW OF ALGEBRA
23
142. −3 < 1/x ≤ 1. We solve the two inequalities separately and take the intersection of the solution sets. First,
−3 < 1/x is clearly true for x > 0. So suppose x < 0. Then −3 < 1/x ⇔ −3x > 1 ⇔ x < −
1
, so for this
3
inequality, the solution set is −∞, −
1
∪ (0, ∞). Now 1/x ≤ 1 is clearly true if x < 0. So suppose x > 0. Then
3
1/x ≤ 1 ⇔ 1 ≤ x, and the solution set here is (−∞, 0) ∪ [1, ∞). Taking the intersection of the two solution
sets gives the ﬁnal solution set: −∞, −
1
∪ [1, ∞).
3
143. C =
5
9
C + 32. So 50 ≤ F ≤ 95 ⇒ 50 ≤
9
9
(F − 32) ⇒ F =
C + 32 ≤ 95 ⇒ 18 ≤
C ≤ 63 ⇒
9
5
5
5
10 ≤ C ≤ 35. So the interval is [10, 35].
144. Since 20 ≤ C ≤ 30 and C =
5
(F − 32), we have 20 ≤
5
(F − 32) ≤ 30 ⇒ 36 ≤ F − 32 ≤ 54 ⇒
9
9
68 ≤ F ≤ 86. So the interval is [68, 86].
145. (a) Let T represent the temperature in degrees Celsius and h the height in km. T = 20 when h = 0 and T decreases
by 10
C for every km (1
C for each 100-m rise). Thus, T = 20 − 10h when 0 ≤ h ≤ 12.
(b) From part (a), T = 20 − 10h ⇒ 10h = 20 − T
h = 2 − T /10. So 0 ≤ h ≤ 5 ⇒
0 ≤ 2 − T /10 ≤ 5 ⇒ −2 ≤ −T /10 ≤ 3 ⇒ −20 ≤ −T ≤ 30 ⇒ 20 ≥ T ≥ −30 ⇒
−30 ≤ T ≤ 20. Thus, the range of temperatures (in
C) to be expected is [−30, 20].
146. The ball will be at least 32 ft above the ground if h ≥ 32 ⇔ 128 + 16t − 16t
2
≥ 32 ⇔
− 16t − 96 ≤ 0 ⇔ 16(t − 3)(t + 2) ≤ 0. t = 3 and t = −2 are endpoints of the interval we’re looking for,
2
16t
and constructing a table gives −2 ≤ t ≤ 3. But t ≥ 0, so the ball will be at least 32 ft above the ground in the time
interval [0, 3].
147. |x + 3| = |2x + 1| ⇔ either x + 3 = 2x + 1 or x + 3 = − (2x + 1). In the ﬁrst case, x = 2, and in the second
case, x + 3 = −2x − 1
. So the solutions are −
and 2.
4
4
3x = −4 ⇔ x = −
3
3
148. |3x + 5| = 1 ⇔ either 3x + 5 = 1 or −1. In the ﬁrst case, 3x = −4 ⇔ x = −
, and in the second case,
4
3
3x = −6 ⇔ x = −2. So the solutions are −2 and −
4
.
3
149. By Property 5 of absolute values, |x| < 3 ⇔ −3 < x < 3, so x ∈ (−3, 3).
150. By Properties 4 and 6 of absolute values, |x| ≥ 3 ⇔ x ≤ −3 or x ≥ 3, so x ∈ (−∞, −3] ∪ [3, ∞).
151. |x − 4| < 1 ⇔ −1 < x − 4 < 1 ⇔ 3 < x < 5, so x ∈ (3, 5).
152. |x − 6| < 0.1 ⇔ −0.1 < x − 6 < 0.1 ⇔ 5.9 < x < 6.1, so x ∈ (5.9, 6.1).
153. |x + 5| ≥ 2 ⇔ x + 5 ≥ 2 or x + 5 ≤ −2 ⇔ x ≥ −3 or x ≤ −7, so x ∈ (−∞, −7] ∪ [−3, ∞).
154. |x + 1| ≥ 3 ⇔ x + 1 ≥ 3 or x + 1 ≤ −3 ⇔ x ≥ 2 or x ≤ −4, so x ∈ (−∞, −4] ∪ [2, ∞).
155. |2x − 3| ≤ 0.4 ⇔ −0.4 ≤ 2x − 3 ≤ 0.4 ⇔ 2.6 ≤ 2x ≤ 3.4 ⇔ 1.3 ≤ x ≤ 1.7, so x ∈ [1.3, 1.7].
156. |5x − 2| < 6 ⇔ −6 < 5x − 2 < 6 ⇔ −4 < 5x < 8 ⇔ −
4
8
, so x ∈ −
4
8
.
< x <
,
5
5
5
5
bc
bc
bc + ac
bc + ac
157. a(bx − c) ≥ bc ⇔ bx − c ≥
⇔ bx ≥
+ c =
⇔ x ≥
a
a
a
ab
c − b
158. ax + b < c ⇔ ax < c − b ⇔ x >
(since a < 0)
a
159. |ab| =
2
2
2
2
2
(ab)
=
a
b
=
a
b
= |a| |b|
160. If 0 < a < b, then a · a < a · b and a · b < b · b [using Rule 3 of Inequalities]. So a
2
2
and hence a
2
2
.
< ab < b
< b