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: 62

Answer

$\left( -2,\dfrac{1}{2} \right)$

Work Step by Step

Using the properties of equality, the given expression, $ -9+|3+4x|\lt-4 ,$ is equivalent to \begin{array}{l}\require{cancel} |3+4x|\lt-4+9 \\\\ |3+4x|\lt5 .\end{array} Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$, then the expression, $ |3+4x|\lt5 ,$ is equivalent to \begin{array}{l}\require{cancel} -5\lt 3+4x\lt5 \\\\ -5-3\lt 3+4x-3\lt5-3 \\\\ -8\lt 4x\lt2 \\\\ -2\lt x\lt\dfrac{1}{2} .\end{array} Hence, the solution set is $ \left( -2,\dfrac{1}{2} \right) .$
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