Chapter 4 - Section 4.3 - Equations and Inequalities Involving Absolute Value - Exercise Set - Page 283: 37

{$-1,15$}

Given: $\dfrac{2x}{3}-2=\dfrac{x}{3}+3$ or,$\dfrac{2x}{3}-\dfrac{x}{3}=3+2$ or, $ x=15$ Now, $\dfrac{2x}{3}-2=-(\dfrac{x}{3}+3)$ or,$\dfrac{2x}{3}-2=-\dfrac{x}{3}-3$ or,$\dfrac{2x}{3}+\dfrac{x}{3}=-3+2$ or, $ x=-1$ Hence, {$-1,15$}