Example 9 Given real numbers $x, y, z$ satisfy $x+y=5$, $z^{2}=x y+y-9$. Then, $x+2 y+3 z=$ $\qquad$
(1995, Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
Solution: The problem provides two equations with three unknowns. Generally speaking, when the number of unknowns exceeds the number of equations, it is impossible to solve for each unknown individually. However, under the condition that $x$, $y$, and $z$ are all real numbers, a solution may still be possible. In fact, from $x+y=5$, we get $x=5-y$, substituting this into the second equation, we have
$$
z^{2}=y(5-y)+y-9=-(y-3)^{2}.
$$

The left side of the equation is a non-negative real number, so the right side must also be a non-negative real number. However, $-(y-3)^{2}$ is a non-positive real number. Therefore, $y-3=0$. Thus, $y=3$. We can then find
$$
\begin{array}{l}
x=5-y=2, z^{2}=0, z=0 . \\
\text { Hence } x+2 y+3 z=2+2 \times 3+3 \times 0=8 .
\end{array}
$$