Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Chapter 1 Test - Page 134: 24


$\left( \dfrac{1}{3},\dfrac{7}{3} \right)$

Work Step by Step

The given inequality, $ |-3x+4|-4\lt-1 ,$ is equivalent to \begin{array}{l}\require{cancel} |-3x+4|\lt-1+4 \\\\ |-3x+4|\lt3 .\end{array} Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the inequality above is equivalent to \begin{array}{l}\require{cancel} -3\lt -3x+4\lt3 .\end{array} Using the properties of inequality, then \begin{array}{l}\require{cancel} -3-4\lt -3x+4-4\lt3-4 \\\\ -7\lt -3x\lt-1 \\\\ \dfrac{-7}{-3}\gt \dfrac{-3x}{-3}\gt\dfrac{-1}{-3} \\\\ \dfrac{7}{3}\gt x\gt\dfrac{1}{3} \\\\ \dfrac{1}{3}\lt x\lt\dfrac{7}{3} .\end{array} Hence, the solution is the interval $ \left( \dfrac{1}{3},\dfrac{7}{3} \right) .$
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