**Unformatted text preview: **3|2x + 1| > −2 analytically, we ﬁrst isolate the absolute value before applying
Theorem 2.3. To that end, we get −3|2x + 1| > −6 or |2x + 1| < 2. Rewriting, we now have
3
−2 < 2x + 1 < 2 so that − 3 < x < 1 . In interval notation, we write − 2 , 1 . Graphically we
2
2
2
1
see the graph of y = 4 − 3|2x + 1| is above y = −2 for x values between − 3 and 2 .
2
y
4
3
2
1 −2 −1 1 2 x −1
−2
−3
−4 3. Rewriting the compound inequality 2 < |x − 1| ≤ 5 as ‘2 < |x − 1| and |x − 1| ≤ 5’ allows
us to solve each piece using Theorem 2.3. The ﬁrst inequality, 2 < |x − 1| can be re-written
as |x − 1| > 2 and so x − 1 < −2 or x − 1 > 2. We get x < −1 or x > 3. Our solution
to the ﬁrst inequality is then (−∞, −1) ∪ (3, ∞). For |x − 1| ≤ 5, we combine results in
Theorems 2.1 and 2.3 to get −5 ≤ x − 1 ≤ 5 so that −4 ≤ x ≤ 6, or [−4, 6]. Our solution to
2 < |x − 1| ≤ 5 is comprised of values of x which satisfy both parts of the inequality, and so we
take what’s called the ‘set theoretic intersection’ of (−∞, −1) ∪...

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