Intermediate Algebra (6th Edition)

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

Chapter 2 - Review - Page 113: 101

Answer

$\text{the interval } \left( -\dfrac{1}{2},2 \right) $

Work Step by Step

Since, for any positive number $c$, $|x|\lt c$ implies $-c \lt x \lt c$, then the solution of the given inequality, $ \left| \dfrac{4x-3}{5} \right|\lt 1 ,$ is \begin{array}{l}\require{cancel} -1 \lt \dfrac{4x-3}{5} \lt 1 \\\\ -1\cdot5 \lt \dfrac{4x-3}{5}\cdot5 \lt 1\cdot5 \\\\ -5 \lt 4x-3 \lt 5 \\\\ -5+3 \lt 4x-3+3 \lt 5+3 \\\\ -2 \lt 4x \lt 8 \\\\ -\dfrac{2}{4} \lt \dfrac{4}{4}x \lt \dfrac{8}{4} \\\\ -\dfrac{1}{2} \lt x \lt 2 .\end{array} In interval notation, the solution is $ \text{the interval } \left( -\dfrac{1}{2},2 \right) .$
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.