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 105: 3

Answer

$(1,5)$

Work Step by Step

Since $|x|\lt c$ implies $-c\lt x\lt c,$ then the solution to the given inequality, $ |x-3|\lt2 ,$ is equivalent to \begin{array}{l}\require{cancel} -2\lt x-3\lt2 \\\\ -2+3\lt x-3+3\lt2+3 \\\\ 1\lt x\lt5 .\end{array} The graph of the solution set is shown below. In interval notation, this is equivalent $ (1,5) .$
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