College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.8 - Solving Absolute Value Equations and Inequalities - 1.8 Exercises - Page 153: 48


Interval: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$

Work Step by Step

$\frac{1}{|2x-3|}\leq 5$ We see that $x\ne\frac{3}{2}$ because otherwise we would get division by 0. $1\leq 5|2x-3|$ $\frac{1}{5}\leq|2x-3|$ We split the absolute value into the positive and negative: $\frac{1}{5}\leq 2x-3$ $\frac{16}{5}\leq 2x$ $\frac{8}{5}\leq x$ $2x-3 \leq-\frac{1}{5}$ $2x\leq\frac{14}{5}$ $ x\leq\frac{7}{5}$ Interval: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$
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.