Intermediate Algebra (12th Edition)

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

Chapter 1 - Summary Exercises - Reviewing a Chapter - Solving Linear and Absolute Value Equations and Inequalities: 35

Answer

$\left( -\infty, -1 \right] \cup \left[ \dfrac{5}{3}, \infty\right]$

Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $ |1-3x|\ge4 ,$ is equivalent to \begin{array}{l}\require{cancel} 1-3x\ge4 \text{ OR } 1-3x\le-4 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 1-3x\ge4 \\\\ -3x\ge4-1 \\\\ -3x\ge3 \\\\ x\le\dfrac{3}{-3} \\\\ x\le-1 \\\\\text{ OR }\\\\ 1-3x\le-4 \\\\ -3x\le-4-1 \\\\ -3x\le-5 \\\\ x\ge\dfrac{-5}{-3} \\\\ x\ge\dfrac{5}{3} .\end{array} Hence, the solution is the interval $ \left( -\infty, -1 \right] \cup \left[ \dfrac{5}{3}, \infty\right] .$
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