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: 73

Answer

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

Work Step by Step

Since $|x|\lt a$ implies $-a\lt x\lt a$, then the statement $ |2x+1|+4\lt7 $ evaluates to \begin{array}{l} |2x+1|\lt3\\ -3\lt2x+1\lt3\\ -4\lt2x\lt2\\ -2\lt x\lt1 .\end{array} Hence, the solution set is $ \text{the interval } \left( -2,1 \right) $.
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