Example 26 Given $-1 \leqslant a+b \leqslant 1,1 \leqslant a-2 b \leqslant 3$. Find the range of $a+3 b$.


Introducing parameters $m, n$, let $a+3b=m(a+b)+n(a-2b)$, then $a+3b=(m+n)a+(m-2n)b$, thus, $\left\{\begin{array}{l}m-2n=3, \\ m+n=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}m=\frac{5}{3}, \\ n=-\frac{2}{3} .\end{array}\right.$ Therefore, $a+3b=\frac{5}{3}(a+b)-\frac{2}{3}(a-2b)$.

From the given conditions,
$$
\begin{array}{l}
-\frac{5}{3} \leqslant \frac{5}{3}(a+b) \leqslant \frac{5}{3} \\
-2 \leqslant-\frac{2}{3}(a-2b) \leqslant-\frac{2}{3} .
\end{array}
$$

Adding them, we get $\quad-\frac{11}{3} \leqslant a+3b \leqslant 1$.