Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 167: 39


The solution is $\Big[-1,-\dfrac{1}{2}\Big]$

Work Step by Step

$\Big|\dfrac{2}{3}x+\dfrac{1}{2}\Big|\le\dfrac{1}{6}$ Solving this absolute value inequality is equivalent to solving the following inequality: $-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}$ $\textbf{Solve the inequality shown above:}$ $-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}$ Multiply the whole inequality by $6$: $6\Big(-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}\Big)$ $-1\le4x+3\le1$ Subtract $3$ from all three parts of the inequality: $-1-3\le4x+3-3\le1-3$ $-4\le4x\le-2$ Divide all three parts of the inequality by $4$: $-\dfrac{4}{4}\le\dfrac{4x}{4}\le-\dfrac{2}{4}$ $-1\le x\le-\dfrac{1}{2}$ Expressing the solution in interval notation: $\Big[-1,-\dfrac{1}{2}\Big]$
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