College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises - Page 166: 117

Answer

$\left[ -6,-3 \right]$

Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$, then the solution to the given inequality, $ |2x+9|\le3 ,$ is \begin{array}{l}\require{cancel} -3\le 2x+9\le3 \\\\ -3-9\le 2x+9-9\le3-9 \\\\ -12\le 2x\le-6 \\\\ -\dfrac{12}{2}\le \dfrac{2x}{2}\le-\dfrac{6}{2} \\\\ -6\le x\le-3 .\end{array} Hence, the solution set is the interval $ \left[ -6,-3 \right] .$
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