Answer
$x\lt\frac{1}{6}$ or $x\gt\frac{13}{6}$
Work Step by Step
We simplify as follows:
$|\frac{2x-3}{2}+\frac{1}{3}|\gt1$
$\frac{2x-3}{2}+\frac{1}{3}\lt-1$ or $\frac{2x-3}{2}+\frac{1}{3}\gt1$
$\frac{2x-3}{2}\lt-\frac{4}{3}$ or $\frac{2x-3}{2}\gt\frac{2}{3}$
$2x-3\lt-\frac{8}{3}$ or $2x-3\gt\frac{4}{3}$
$2x\lt\frac{1}{3}$ or $2x\gt\frac{13}{3}$
$x\lt\frac{1}{6}$ or $x\gt\frac{13}{6}$
