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: 59

Answer

$ \left( -\infty, -12 \right)\cup\left( 0,\infty \right) $
1502205798

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a\text{ or } x\lt-a$, then the expression, $ \left| \dfrac{x+6}{3} \right|\gt2 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{x+6}{3}\gt2 \\\\ x+6\gt2(3) \\\\ x+6\gt6 \\\\ x\gt6-6 \\\\ x\gt0 ,\\\\\text{ OR }\\\\ \dfrac{x+6}{3}\lt-2 \\\\ x+6\lt-2(3) \\\\ x+6\lt-6 \\\\ x\lt-6-6 \\\\ x\lt-12 .\end{array} Hence, the solution set is $ \left( -\infty, -12 \right)\cup\left( 0,\infty \right) .$
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