解不等式 $x - 3 \le \dfrac{5 - x}{3} < \dfrac{x + 1}{2}$。

1. Split the compound inequality into two inequalities.

$x-3 \le \frac{5-x}{3}\quad \text{and}\quad \frac{5-x}{3}<\frac{x+1}{2}$

2. Solve the left-hand inequality.

$x-3 \le \frac{5-x}{3}$
Multiply by $3$:
$3x-9 \le 5-x$
$4x \le 14$
$x \le \frac{7}{2}$

3. Solve the right-hand inequality and combine both results.

$\frac{5-x}{3}<\frac{x+1}{2}$
Multiply by $6$:
$2(5-x)<3(x+1)$
$10-2x<3x+3$
$7<5x$
$x>\frac{7}{5}$

Combining $x\le \frac{7}{2}$ and $x>\frac{7}{5}$ gives
$\frac{7}{5}<x\le \frac{7}{2}.$