Solve |3−4x|≥9

As, $|3-4 x| \geq 9$

$\Rightarrow(3-4 x) \leq-9$ or $(3-4 x) \geq 9 \quad($ As, $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a)$

$\Rightarrow-4 x \leq-9-3$ or $-4 x \geq 9-3$

$\Rightarrow-4 x \leq-12$ or $-4 x \geq 6$

$\Rightarrow x \geq \frac{-12}{-4}$ or $x \leq \frac{6}{-4}$

$\Rightarrow x \geq 3$ or $x \leq \frac{-3}{2}$

$\therefore x \in\left(-\infty, \frac{-3}{2}\right] \cup[3, \infty)$