Answer
The solution is $(-\infty,0)\cup(6,\infty)$
Work Step by Step
$4|x-3|\gt12$
Take $4$ to divide the right side:
$|x-3|\gt\dfrac{12}{4}$
$|x-3|\gt3$
Solving this absolute value inequality is equivalent to solving two separate inequalities, which are:
$x-3\gt3$ and $x-3\lt-3$
$\textbf{Solve the first inequality:}$
$x-3\gt3$
Take $3$ to the right side:
$x\gt3+3$
$x\gt6$
$\textbf{Solve the second inequality:}$
$x-3\lt-3$
Take $3$ to the right side:
$x\lt-3+3$
$x\lt0$
Expressing the solution in interval notation:
$(-\infty,0)\cup(6,\infty)$
