Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 79


$x \lt -\dfrac{43}{24} \text{ or } x \gt \dfrac{9}{8}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $ \left| \dfrac{1+3x}{5} \right| \gt \dfrac{7}{8} ,$ use the definition of a greater than (greater than or equal to) absolute value inequality and solve each resulting inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1+3x}{5} \gt \dfrac{7}{8} \\\\\text{OR}\\\\ \dfrac{1+3x}{5} \lt -\dfrac{7}{8} .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} \dfrac{1+3x}{5} \gt \dfrac{7}{8} \\\\ 40\cdot\dfrac{1+3x}{5} \gt 40\cdot\dfrac{7}{8} \\\\ 8\cdot(1+3x) \gt 5\cdot7 \\\\ 8+24x \gt 35 \\\\ 24x \gt 35-8 \\\\ 24x \gt 27 \\\\ x \gt \dfrac{27}{24} \\\\ x \gt \dfrac{9}{8} \\\\\text{OR}\\\\ \dfrac{1+3x}{5} \lt -\dfrac{7}{8} \\\\ 40\cdot\dfrac{1+3x}{5} \lt 40\cdot\left( -\dfrac{7}{8} \right) \\\\ 8\cdot(1+3x) \lt 5\cdot\left( -7 \right) \\\\ 8+24x \lt -35 \\\\ 24x \lt -35-8 \\\\ 24x \lt -43 \\\\ x \lt -\dfrac{43}{24} .\end{array} Hence, the solution set is $ x \lt -\dfrac{43}{24} \text{ or } x \gt \dfrac{9}{8} .$
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