Example 12 Given that $x, y, z$ are 3 non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $s=2x+y-z$, then what is the sum of the maximum and minimum values of $s$?
(1996, Tianjin Junior High School Mathematics Competition)
The given equations and the expression for $s$ can be viewed as a system of equations in $x$, $y$, and $z$:
$$
\left\{\begin{array}{l}
3 x+2 y+z=5, \\
x+y-z=2, \\
2 x+y-z=s .
\end{array}\right.
$$

Solving this system, we get $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
Since $x$, $y$, and $z$ are non-negative rational numbers, we have $\left\{\begin{array}{l}s-2 \geqslant 0, \\ 5-\frac{4 s}{3} \geqslant 0, \\ -\frac{s}{3}+1 \geqslant 0 .\end{array} \quad\right.$ Solving these inequalities, we get $\left\{\begin{array}{l}s \geqslant 2, \\ s \leqslant \frac{15}{4}, \\ s \leqslant 3 .\end{array}\right.$
Therefore, the sum of the maximum and minimum values of $s$ is $3+2=$
5.