## Precalculus (6th Edition) Blitzer

$(-\infty,-2)\cup(6,\infty)$
Rewrite as $|2-x| \gt 4$ ... also, $|a-b|=|b-a|$ $|x-2| \gt 4$ ... $|u| \gt c$ is equivalent to ($u\lt -c$) or ($u\gt c$) $\begin{array}{lllll} x-2\lt-4 & /+2 & ...or... & x-2\gt 4 & /+2\\ x\lt-2 & & & x \gt 6 & \\ x\in(-\infty,-2) & & or & x\in(6,\infty) & \\ & & & & \end{array}$ Solution set: $(-\infty,-2)\cup(6,\infty)$