Answer
$0\lt x\lt3$
See graph
Work Step by Step
Apply absolute rule:
If $|u|\lt a,~a\gt 0$, then $-a\lt u\lt a$.
Take $u=1-\frac{2x}{3}$ and $a=1$.
$$-1\lt1-\frac{2x}{3}\lt1$$
$$-1\lt1-\frac{2x}{3}\text{ and }1-\frac{2x}{3}\lt1$$
$$-1-1\lt1-\frac{2x}{3}-1\text{ and }1-\frac{2x}{3}-1\lt1-1$$
$$-2\lt-\frac{2x}{3}\text{ and }-\frac{2x}{3}\lt0$$
$$-2\left(-\frac{3}{2}\right)\gt-\frac{2x}{3}\left(-\frac{3}{2}\right) \text{ and }-\frac{2x}{3}\left(-\frac{3}{2}\right) \gt0\left(-\frac{3}{2}\right)$$
$$3\gt x\text{ and }x\gt0$$
Combining:
$$0\lt x\lt3$$
The graph of the solution set is as shown.
