Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Chapter 1 Test: 22

Answer

$\left( -\infty, -\dfrac{7}{6} \right) \cup \left( \dfrac{17}{6}, \infty\right)$

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $ |5-6x|\gt12 ,$ is equivalent to \begin{array}{l}\require{cancel} 5-6x\gt12 \text{ OR } 5-6x\lt-12 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 5-6x\gt12 \\\\ -6x\gt12-5 \\\\ -6x\gt7 \\\\ x\lt\dfrac{7}{-6} \\\\ x\lt-\dfrac{7}{6} \\\\\text{ OR }\\\\ 5-6x\lt-12 \\\\ -6x\lt-12-5 \\\\ -6x\lt-17 \\\\ x\gt\dfrac{-17}{-6} \\\\ x\gt\dfrac{17}{6} .\end{array} Hence, the solution is the interval $ \left( -\infty, -\dfrac{7}{6} \right) \cup \left( \dfrac{17}{6}, \infty\right) .$
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