Intermediate Algebra (6th Edition)

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

Chapter 2 - Review - Page 113: 79

Answer

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

Work Step by Step

Using the properties of equality, the given expression, $ \left| 3x \right|-8\gt1 ,$ is equivalent to \begin{array}{l}\require{cancel} \left| 3x \right|\gt1+8 \\\\ \left| 3x \right|\gt9 .\end{array} Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a\text{ or } x\lt-a$, then the expression, $ \left| 3x \right|\gt9 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x\gt9 \\\\ x\gt\dfrac{9}{3} \\\\ x\gt3 ,\\\\\text{ OR }\\\\ 3x\lt-9 \\\\ x\lt-\dfrac{9}{3} \\\\ x\lt-3 .\end{array} Hence, the solution set is $ \left( -\infty, -3 \right)\cup\left( 3,\infty \right) .$
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