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 599: 70


$a\le-\dfrac{13}{2} \text{ or } a\ge\dfrac{3}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Solve the given inequality, $ |2a+5|+1\ge9 ,$ by isolating first the absolute value expression. Then use the definition of a greater than (greater than or equal to) absolute value 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 $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given expression is equivalent to \begin{array}{l}\require{cancel} |2a+5|+1\ge9 \\\\ |2a+5|\ge9-1 \\\\ |2a+5|\ge8 .\end{array} 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 given inequality is equivalent to \begin{array}{l}\require{cancel} 2a+5\ge8 \\\\\text{OR}\\\\ 2a+5\le-8 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2a+5\ge8 \\\\ 2a\ge8-5 \\\\ 2a\ge3 \\\\ a\ge\dfrac{3}{2} \\\\\text{OR}\\\\ 2a+5\le-8 \\\\ 2a\le-8-5 \\\\ 2a\le-13 \\\\ a\le-\dfrac{13}{2} .\end{array} Hence, the solution set is $ a\le-\dfrac{13}{2} \text{ or } a\ge\dfrac{3}{2} .$
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