## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 36: 2

#### Answer

(a) $x\in(1,9)$ (b) $x\in[-1,-1/5]$.

#### Work Step by Step

(a) $$|x-5|\lt 4\Longrightarrow 5-4\lt x\lt 4+5 \Longrightarrow 1\lt x\lt 9.$$ So, $x\in(1,9)$ (b) $$|5x+3|\leq 2\Longrightarrow -3-2\leq 5x\leq 2-3 \\ \Longrightarrow -5\leq 5x\leq -1\Longrightarrow -1\leq x\leq -1/5.$$ So, $x\in[-1,-1/5]$.

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