Answer
$(-\infty,1/3)\cup(1,\infty)$
Work Step by Step
Rewrite as
$|2-3x| \gt 1$
... also, $|a-b|=|b-a|$
$|3x-2| \gt 1$
... $|u| \gt c $ is equivalent to ($ u\lt -c $) or ($ u\gt c $)
$ \begin{array}{lllll}
3x-2\lt-1 & /+2 & ...or... & 3x-2\gt 1 & /+2\\
3x\lt 1 & /\div 3 & & 3x\gt 3 & /\div 3\\
x\lt 1/3 & & & x \gt 1 & \\
x\in(-\infty,1/3) & & or & x\in(1,\infty) & \\
& & & &
\end{array} $
Solution set: $(-\infty,1/3)\cup(1,\infty)$
