Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.9 - Linear Inequalities and Absolute Value Inequalities - Exercise Set - Page 138: 90


solution set: $\mathbb{R}$

Work Step by Step

Rewrite as $|x-\displaystyle \frac{11}{3}|+\frac{7}{3} \gt 1\qquad $... $/\times 3$ $ 3|x-\displaystyle \frac{11}{3}|+7 \gt 3\qquad $... $/-7$ $|3x-11| \gt -4 $ Whatever x is, the absolute value is always going to be greater than a negative number, so all real x values are solutions. Solution set: $(-\infty,\infty)$
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