Answer
$(-\infty,-1)\cup(\frac{7}{3},\infty)$
See graph.
Work Step by Step
Step 1. Remove the absolute value sign and transform the original inequality:
$1-|2-3x|\lt-4\Longrightarrow |3x-2|\gt5 \Longrightarrow 3x-2\lt-5\ or\ 3x-2\gt5 \Longrightarrow x\lt-1\ or\ x\gt\frac{7}{3}$, using interval form $(-\infty,-1)\cup(\frac{7}{3},\infty)$
Step 2. See graph.
