Intermediate Algebra (6th Edition)

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

Chapter 2 - Test - Page 114: 17

Answer

$\text{the interval } \left( -\infty, -2 \right) \cup \left( \dfrac{4}{3},\infty \right) $

Work Step by Step

Since for any nonnegative $c$, $|x|\gt c$ implies $x\gt c$ or $x\lt -c$, then the solution to the given inequality, $ |3x+1|\gt5 ,$ is \begin{array}{l}\require{cancel} 3x+1\gt5 \\\\ 3x\gt5-1 \\\\ 3x\gt4 \\\\ x\gt\dfrac{4}{3} \\\\\text{ OR }\\\\ 3x+1\lt-5 \\\\ 3x\lt-5-1 \\\\ 3x\lt-6 \\\\ x\lt-\dfrac{6}{3} \\\\ x\lt-2 .\end{array} In interval notation, the solution is $ \text{the interval } \left( -\infty, -2 \right) \cup \left( \dfrac{4}{3},\infty \right) .$
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