Let $a$ be a given real number, solve the system of inequalities about $x$
$$
a \cdot \frac{1-x}{2} \leqslant \frac{x+2}{3} \leqslant a \cdot \frac{3-x}{4} \text {. }
$$
Solution: The original system of inequalities is equivalent to
$$
\left\{\begin{array}{l}
(3 a+2) x \geqslant 3 a-4, \\
(3 a+4) x \leqslant 9 a-8 .
\end{array}\right.
$$

We classify and discuss based on the coefficient of $x$ in each inequality.
(1) If $3 a+2>0$, i.e., $a>-\frac{2}{3}$, then $3 a+4>0$. Thus, the original system of inequalities becomes
$$
\frac{3 a-4}{3 a+2} \leqslant x \leqslant \frac{9 a-8}{3 a+4} \text {. }
$$

Notice that
$$
\begin{array}{l}
\frac{3 a-4}{3 a+2} \leqslant \frac{9 a-8}{3 a+4} \\
\Leftrightarrow a\left(a-\frac{1}{3}\right) \geqslant 0 \Leftrightarrow a \leqslant 0 \text { or } a \geqslant \frac{1}{3} .
\end{array}
$$

Therefore, when $-\frac{2}{3}<a \leqslant 0$ or $a \geqslant \frac{1}{3}$, the solution to the original system of inequalities is $\frac{3 a-4}{3 a+2} \leqslant x \leqslant \frac{9 a-8}{3 a+4}$.

(2) If $3 a+2<0$, i.e., $-\frac{4}{3}<a<-\frac{2}{3}$, then the original system of inequalities becomes
$$
\left\{\begin{array}{l}
x \leqslant \frac{3 a-4}{3 a+2}, \\
x \leqslant \frac{9 a-8}{3 a+4} .
\end{array}\right.
$$

Notice that
$$
\begin{array}{l}
\frac{9 a-8}{3 a+4} \geqslant \frac{3 a-4}{3 a+2} \\
\Leftrightarrow(9 a-8)(3 a+2) \leqslant(3 a+4)(3 a-4) \\
\Leftrightarrow 0 \leqslant a \leqslant \frac{1}{3} .
\end{array}
$$

Thus, when $-\frac{4}{3}<a<-\frac{2}{3}$, it must be that $\frac{9 a-8}{3 a+4} \leqslant \frac{3 a-4}{3 a+2}$. Therefore, the solution to the original inequality is $x \leqslant \frac{9 a-8}{3 a+4}$.

In summary, when $a \leqslant-\frac{4}{3}$ or $0<a<\frac{1}{3}$, the original system of inequalities has no solution; when $-\frac{4}{3}<a \leqslant-\frac{2}{3}$, the solution to the original system of inequalities is $x \leqslant \frac{9 a-8}{3 a+4}$; when $-\frac{2}{3}<a \leqslant 0$ or $a \geqslant \frac{1}{3}$, the solution to the original system of inequalities is $\frac{3 a-4}{3 a+2} \leqslant x \leqslant \frac{9 a-8}{3 a+4}$.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)