Intermediate Algebra (12th Edition)

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

Chapter 1 - Mixed Review Exercises: 13

Answer

$\left( -\infty, -\dfrac{13}{5} \right) \cup \left( 3, \infty\right)$

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $ |5x-1|\gt14 ,$ is equivalent to \begin{array}{l}\require{cancel} 5x-1\gt14 \text{ OR } 5x-1\lt-14 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 5x-1\gt14 \\\\ 5x\gt14+1 \\\\ 5x\gt15 \\\\ x\gt\dfrac{15}{5} \\\\ x\gt3 \\\\\text{ OR }\\\\ 5x-1\lt-14 \\\\ 5x\lt-14+1 \\\\ 5x\lt-13 \\\\ x\lt-\dfrac{13}{5} .\end{array} Hence, the solution is the interval $ \left( -\infty, -\dfrac{13}{5} \right) \cup \left( 3, \infty\right) .$
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