Answer
$(-\infty,-3)\cup(5,\infty)$
Work Step by Step
$-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: $(-\infty,-3)\cup(5,\infty)$
