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 - Review Exercises: Chapter 9 - Page 623: 33

Answer

$-\dfrac{7}{2}\lt a \lt 2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |4a+3|\lt11 ,$ use the definition of a less than (less than or equal to) absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -11\lt 4a+3 \lt11 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -11\lt 4a+3 \lt11 \\\\ -11-3\lt 4a+3-3 \lt11-3 \\\\ -14\lt 4a \lt 8 \\\\ -\dfrac{14}{4}\lt \dfrac{4a}{4} \lt \dfrac{8}{4} \\\\ -\dfrac{7}{2}\lt a \lt 2 .\end{array} Hence, the solution set $ -\dfrac{7}{2}\lt a \lt 2 .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.