## Precalculus (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 168: 50

#### Answer

The solution is $\Big(-\dfrac{5}{3},\dfrac{4}{3}\Big)$

#### Work Step by Step

$\Big|2x+\dfrac{1}{3}\Big|+1\lt4$ Take $1$ to the right side of the inequality: $\Big|2x+\dfrac{1}{3}\Big|\lt4-1$ $\Big|2x+\dfrac{1}{3}\Big|\lt3$ Solving this absolute value inequality is equivalent to solving the following inequality: $-3\lt2x+\dfrac{1}{3}\lt3$ Multiply the whole inequality by $3$: $3\Big(-3\lt2x+\dfrac{1}{3}\lt3\Big)$ $-9\lt6x+1\lt9$ Subtract $1$ from all three parts of the inequality: $-9-1\lt6x+1-1\lt9-1$ $-10\lt6x\lt8$ Divide all three parts of the inequality by $6$: $-\dfrac{10}{6}\lt\dfrac{6x}{6}\lt\dfrac{8}{6}$ $-\dfrac{5}{3}\lt x\lt\dfrac{4}{3}$ Expressing the solution in interval notation: $\Big(-\dfrac{5}{3},\dfrac{4}{3}\Big)$

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