Algebra Rules Review Sheet (Page 23 of 23) in pdf

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REVIEW OF ALGEBRA

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 < −

, so for this

inequality, the solution set is −∞, −

∪ (0, ∞). Now 1/x ≤ 1 is clearly true if x < 0. So suppose x > 0. Then

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, ∞).

143. C =

C + 32. So 50 ≤ F ≤ 95 ⇒ 50 ≤

(F − 32) ⇒ F =

C + 32 ≤ 95 ⇒ 18 ≤

C ≤ 63 ⇒

10 ≤ C ≤ 35. So the interval is [10, 35].

144. Since 20 ≤ C ≤ 30 and C =

(F − 32), we have 20 ≤

(F − 32) ≤ 30 ⇒ 36 ≤ F − 32 ≤ 54 ⇒

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

≥ 32 ⇔

− 16t − 96 ≤ 0 ⇔ 16(t − 3)(t + 2) ≤ 0. t = 3 and t = −2 are endpoints of the interval we’re looking for,

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.

⇔

3x = −4 ⇔ x = −

148. |3x + 5| = 1 ⇔ either 3x + 5 = 1 or −1. In the ﬁrst case, 3x = −4 ⇔ x = −

, and in the second case,

3x = −6 ⇔ x = −2. So the solutions are −2 and −

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 ⇔ −

, so x ∈ −

< x <

bc + ac

157. a(bx − c) ≥ bc ⇔ bx − c ≥

⇔ bx ≥

+ c =

⇔ x ≥

c − b

158. ax + b < c ⇔ ax < c − b ⇔ x >

(since a < 0)

√

159. |ab| =

(ab)

= |a| |b|

160. If 0 < a < b, then a · a < a · b and a · b < b · b [using Rule 3 of Inequalities]. So a

and hence a

< ab < b

< b

Algebra Rules Review Sheet Page 23

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