Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.7 - Absolute Value Inequalities - Exercise Set - Page 106: 58


The solution is $\Big[-\dfrac{8}{7},2\Big]$

Work Step by Step

$|7x-3|-1\le10$ Take $1$ to the right side: $|7x-3|\le10+1$ $|7x-3|\le11$ According to an absolute value inequality property, $|x|\le a$ is equivalent to $-a\le x\le a$. Then, in this particular case, solving $|7x-3|\le11$ is equivalent to solving $-11\le7x-3\le11$ $-11\le7x-3\le11$ Add $3$ on all parts of the inequality: $-11+3\le7x-3+3\le11+3$ $-8\le7x\le14$ Divide all parts of the inequality by $7$: $-\dfrac{8}{7}\le\dfrac{7x}{7}\le\dfrac{14}{7}$ $-\dfrac{8}{7}\le x\le2$ Expressing the solution in interval notation: $\Big[-\dfrac{8}{7},2\Big]$
Small 1510077190
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.