## Precalculus (6th Edition) Blitzer

$[0,8]$
$|2-\displaystyle \frac{x}{2}|-1\leq 1 \qquad$... $/+1$ $|2-\displaystyle \frac{x}{2}|\leq 2 \qquad$... $/\times 2$ $|4-x|\leq 4 \qquad$... apply $|a-b|=|b-a|$ $|x-4|\leq 4$ ... $|u| \leq c$ is equivalent to $\quad -c \leq u \leq c.$ $-4 \leq x-4 \leq 4 \qquad$... $/+4$ $0 \leq x \leq 8$ Borders are included it the interval. Solution set: $[0,8]$