Algebra Rules Review Sheet Page 21
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REVIEW OF ALGEBRA
21
115. False. Using Law 3 of Exponents, (x
3
4
3 · 4
12
7
.
)
= x
= x
6 = x
116. True.
117. |5 − 23| = |−18| = 18
118. |π − 2| = π − 2 because π − 2 > 0.
√
√
√
√
119.
5 because
5 − 5 = −
5 − 5 = 5 −
5 − 5 < 0.
120. |−2| − |−3| = |2 − 3| = |−1| = 1
121. If x < 2, x − 2 < 0, so |x − 2| = − (x − 2) = 2 − x.
122. If x > 2, x − 2 > 0, so |x − 2| = x − 2.
if x + 1 ≥ 0
if x ≥ −1
x + 1
x + 1
123. |x + 1| =
=
if x + 1 < 0
if x < −1
−(x + 1)
−x − 1
if 2x − 1 ≥ 0
if x ≥
1
2x − 1
2x − 1
124. |2x − 1| =
2
=
if 2x − 1 < 0
if x <
1
−(2x − 1)
1 − 2x
2
125. x
2
2
+ 1 (since x
2
+ 1 ≥ 0 for all x).
+ 1 = x
√
126. Determine when 1 − 2x
2
2
2
1
1
1
2
< 0 ⇔ 1 < 2x
⇔ x
>
⇔
x
>
⇔ |x| >
⇔
2
2
2
2
if −
1
1
1 − 2x
≤ x ≤
√
√
2
2
or x >
. Thus, 1 − 2x
1
1
2
x < −
=
√
√
2
2
2
− 1 if x < −
1
or x >
1
2x
√
√
2
2
127. 2x + 7 > 3 ⇔ 2x > −4 ⇔ x > −2, so x ∈ (−2, ∞).
128. 4 − 3x ≥ 6 ⇔ −3x ≥ 2 ⇔ x ≤ −
2
, so x ∈ −∞, −
2
.
3
3
129. 1 − x ≤ 2 ⇔ −x ≤ 1 ⇔ x ≥ −1, so x ∈ [−1, ∞).
130. 1 + 5x > 5 − 3x ⇔ 8x > 4 ⇔ x >
, so x ∈
, ∞ .
1
1
2
2
131. 0 ≤ 1 − x < 1 ⇔ −1 ≤ −x < 0 ⇔ 1 ≥ x > 0, so x ∈ (0, 1].
132. 1 < 3x + 4 ≤ 16 ⇔ −3 < 3x ≤ 12 ⇔ −1 < x ≤ 4, so
x ∈ (−1, 4].
133. (x − 1)(x − 2) > 0.
Case 1: (both factors are positive, so their product is positive)
x − 1 > 0 ⇔ x > 1, and x − 2 > 0 ⇔ x > 2, so x ∈ (2, ∞).
Case 2: (both factors are negative, so their product is positive)
x − 1 < 0 ⇔ x < 1, and x − 2 < 0 ⇔ x < 2, so x ∈ (−∞, 1).
Thus, the solution set is (−∞, 1) ∪ (2, ∞).
134. x
2
2
− 2x − 8 < 0 ⇔ (x − 4)(x + 2) < 0. Case 1: x > 4 and x < −2, which is impossible.
< 2x + 8 ⇔ x
Case 2: x < 4 and x > −2. Thus, the solution set is (−2, 4).
√
√
√
√
135. x
2
2
3 < 0. Case 1: x >
3 and x < −
3, which is impossible.
< 3 ⇔ x
− 3 < 0 ⇔
x −
3
x +
√
√
√
√
Case 2: x <
3 and x > −
3. Thus, the solution set is −
3 .
3,
√
√
√
Another method: x
2
< 3 ⇔ |x| <
3 ⇔ −
3 < x <
3.
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