## Algebra 2 (1st Edition)

$x\gt\frac{c-b}{a}$ or $x\lt\frac{-c-b}{a}$.
$|ax+b|\gt c$ implies $ax+b\gt c\\ax\gt c-b\\x\gt\frac{c-b}{a}$ or $ax+b\lt -c\\ax\lt -c-b\\x\lt\frac{-c-b}{a}$. Thus $x\gt\frac{c-b}{a}$ or $x\lt\frac{-c-b}{a}$.