Answer
$\frac{1}{5}\leq x\leq\frac{7}{5}$
See graph
Work Step by Step
$$3|4-5x|\leq9$$
$$\frac{3|4-5x|}{3}\leq\frac{9}{3}$$
$$|4-5x|\leq3$$
Apply absolute rule:
If $|u|\leq a,~a\gt 0$, then $-a\leq u\leq a$.
Take $u=4-5x$ and $a=3$.
$$-3\leq4-5x\leq3$$
$$-3\leq4-5x\text{ and }4-5x\leq3$$
$$-3-4\leq4-5x-4\text{ and }4-5x-4\leq3-4$$
$$-7\leq-5x\text{ and }-5x\leq-1$$
$$\frac{-7}{-5}\geq\frac{-5x}{-5}\text{ and }\frac{-5x}{-5}\geq\frac{-1}{-5}$$
$$\frac{7}{5}\geq x\text{ and }x\geq \frac{1}{5}$$
$$x\leq\frac{7}{5}\text{ and }x\geq \frac{1}{5}$$
Combining:
$$\frac{1}{5}\leq x\leq\frac{7}{5}$$
The graph of the solution set is as shown.
