# Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises - Page 154: 32

$\left( -\dfrac{8}{5},\dfrac{2}{5} \right)$

#### Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x \lt a$ then the solution to the given inequality, $\left| \dfrac{3}{5}+x \right|\lt 1 ,$ is \begin{array}{l}\require{cancel} -1\lt \dfrac{3}{5}+x \lt 1 \\\\ -1-\dfrac{3}{5}\lt -\dfrac{3}{5}+\dfrac{3}{5}+x \lt 1-\dfrac{3}{5} \\\\ -\dfrac{8}{5}\lt x \lt \dfrac{2}{5} .\end{array} Hence, the solution set is the interval $\left( -\dfrac{8}{5},\dfrac{2}{5} \right) .$

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