Answer
$(-\infty,-8)\cup(16,\infty)$.
The graph is shown below.
Work Step by Step
The given expression is
$\Rightarrow \left | 3-\frac{3x}{4}\right |\gt9$
Rewrite the inequality without absolute value bars.
$\Rightarrow 3-\frac{3x}{4}\lt-9$ or $3-\frac{3x}{4}\gt9$
Solve each inequality separately.
Subtract $3$ from all parts.
$\Rightarrow 3-\frac{3x}{4}-3\lt-9-3$ or $3-\frac{3x}{4}-3\gt9-3$
Simplify
$\Rightarrow -\frac{3x}{4}\lt-12$ or $-\frac{3x}{4}\gt6$
Multiply all parts by $-4$ and change the sense of the inequality.
$\Rightarrow -\frac{3x}{4}(-4)\gt-12(-4)$ or $-\frac{3x}{4}(-4)\lt6(-4)$
Simplify.
$\Rightarrow 3x\gt48$ or $3x\lt-24$
Divide all parts by $3$.
$\Rightarrow \frac{3x}{3}\gt\frac{48}{3}$ or $\frac{3x}{3}\lt\frac{-24}{3}$
Simplify.
$\Rightarrow x\gt16$ or $x\lt-8$
The solution set is less than $-8$ or greater than $16$.
The interval notation is
$(-\infty,-8)\cup(16,\infty)$.
