Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 168: 47


The solution is $\Big(-\dfrac{4}{3},\dfrac{2}{3}\Big)$

Work Step by Step

$|3x+1|-1\lt2$ Take $1$ to the right side of the inequality: $|3x+1|\lt2+1$ $|3x+1|\lt3$ Solving this absolute value inequality is equivalent to solving the following inequality: $-3\lt3x+1\lt3$ $\textbf{Solve the inequality shown above:}$ $-3\lt3x+1\lt3$ Subtract $1$ from all three parts of the inequality: $-3-1\lt3x+1-1\lt3-1$ $-4\lt3x\lt2$ Divide all three parts of the inequality by $3$: $-\dfrac{4}{3}\lt\dfrac{3x}{3}\lt\dfrac{2}{3}$ $-\dfrac{4}{3}\lt x\lt\dfrac{2}{3}$ Expressing the solution in interval notation: $\Big(-\dfrac{4}{3},\dfrac{2}{3}\Big)$
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