## Precalculus (6th Edition) Blitzer

$(-\infty,-3)\cup(5,\infty)$
$-4|1-x| \lt -16\qquad$ ... divide with $(-$4$)$ ... $\div$(negative) $\Rightarrow$ change direction $|1-x|\gt 4$ ... $|a-b|=|b-a|$ $|x-1|\gt 4$ ... $|u| \gt c$ is equivalent to ($u\lt -c$) or ($u\gt c$) $\begin{array}{lllll} x-1\lt-4 & /+1 & ...or... & x-1\gt 4 & /+1\\ x\lt-3 & & & x\gt 5 & \\ x\in(-\infty,-3) & & or & x\in(5,\infty) & \\ & & & & \end{array}$ Solution set: $(-\infty,-3)\cup(5,\infty)$