Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.7 - Absolute Value Inequalities - Exercise Set - Page 106: 51

Answer

$\left( -\infty,-3 \right)\cup(0,\infty)$

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ OR $x\lt -a$, then the given inequality, $ \left| \dfrac{2}{3}x+1 \right|\gt1 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}x+1\gt1 \\\\ \dfrac{2}{3}x\gt1-1 \\\\ \dfrac{2}{3}x\gt0 \\\\ \dfrac{3}{2}\cdot\dfrac{2}{3}x\gt0\cdot\dfrac{3}{2} \\\\ x\gt0 ,\\\\\text{OR}\\\\ \dfrac{2}{3}x+1\lt-1 \\\\ \dfrac{2}{3}x\lt-1-1 \\\\ \dfrac{2}{3}x\lt-2 \\\\ \dfrac{3}{2}\cdot\dfrac{2}{3}x\lt-2\cdot\dfrac{3}{2} \\\\ x\lt-3 .\end{array} Hence, the solution set is $ \left( -\infty,-3 \right)\cup(0,\infty) .$ See graph below.
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