Example 8 Given 3 non-negative numbers $a, b, c$ satisfying $3a+2b+c=5$ and $2a+b-3c=1$. If $m=$ $3a+b-7c$, then the minimum value of $m$ is $\qquad$, and the maximum value of $m$ is $\qquad$
(14th Junior High School Mathematics Competition in Jiangsu Province)
Solution: From the equations $3 a+2 b+c=5$ and $2 a+b-3 c=1$, we solve to get
$$
a=7 c-3, b=7-11 c .
$$

Substituting into $m=3 a+b-7 c$, we get
$$
m=3 c-2 \text{. }
$$

Since $a \geqslant 0, b \geqslant 0, c \geqslant 0$, we have
$$
\left\{\begin{array}{l}
a=7 c-3 \geqslant 0, \\
b=7-11 c \geqslant 0, \\
c \geqslant 0 .
\end{array}\right.
$$

Solving the system of inequalities, we get
$$
\frac{3}{7} \leqslant c \leqslant \frac{7}{11} \text{. }
$$

From (1) and (2), we get
$$
-\frac{5}{7} \leqslant m \leqslant-\frac{1}{11} \text{. }
$$

Therefore, the minimum value of $m$ is $-\frac{5}{7}$, and the maximum value is $-\frac{1}{11}$.