## Precalculus (6th Edition) Blitzer

$[-2,-\displaystyle \frac{2}{9}]$
4 times a number is subtracted from 5 can be written as: $5-4x$ The absolute value of the difference $|5-4x|$ is at most 13: $|5-4x|\leq 13$ $|5-4x|=|4x-5| \qquad$, because $|a-b|=|b-a|$ So, we solve $|4x-5|\leq 13$ $|u| \leq c$ is equivalent to $-c \leq u \leq c.$ $-13\leq 4x-5\leq 13\qquad.../+5$ $-8\leq 4x\leq 18 \qquad.../\div 4$ $-2\displaystyle \leq x\leq\frac{9}{2}$ Borders are included: Solution set: $[-2,-\displaystyle \frac{2}{9}]$