9. Given that $x, y, z$ are three non-negative rational numbers, and satisfy $3 x$ $+2 y+z=5, x+y-z=2$. If $S=2 x+y-z$, then what is the sum of the maximum and minimum values of $S$?
9. $\left\{\begin{array}{l}3 x+y+z=5, \\ x+y-z=2, \\ 2 x+y-z=S .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}x=S-2, \\ y=\frac{15-4 S}{3}, \\ z=\frac{3-S}{3} .\end{array}\right.$ Since $x, y, z$ are all non-negative, we have
$$
\left\{\begin{array} { l } 
{ S - 2 \geqslant 0 , } \\
{ \frac { 1 5 - 4 S } { 3 } \geqslant 0 , } \\
{ \frac { 3 - S } { 3 } \geqslant 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
S \geqslant 2, \\
S \leqslant \frac{15}{4}, \\
S \leqslant 3 .
\end{array}\right.\right.
$$

Therefore, $2 \leqslant S \leqslant 3$. The sum of the maximum and minimum values of $S$ is 5.