Three, given $y=y_{1}-$
$y_{2}, y_{1}$ is inversely proportional to $x$,
$y_{2}$ is directly proportional to $x-2$, when
$x=1$, $y=-1$; when $x$
$=-3$, $y_{1}$ is 21 less than $y_{2}$.
Find the corresponding value of $y$ when $x=-5$. (10 points)
Three, Solve $\because y_{1}$ is inversely proportional to $x$, let $y_{1}=\frac{k_{1}}{x}-$, $y_{2}$ is directly proportional to $x-2$, let $y_{2}=k_{2}(x-2)$. Thus, $y=\frac{k_{1}}{x}--k_{2}(x-2)$.
Substitute $x=1, y=-1$ into the above equation to get $k_{1}+k_{2}=-1$.
When $x=-3$, $y_{1}$ is 21 less than $y_{2}$,

we have $-\frac{k_{1}}{3}+21=-5 k_{2}$,
which gives $k_{1}-15 k_{2}=63$.
Solve the system of equations: $\left\{\begin{array}{l}k_{1}+k_{2}=-1, \\ k_{1}-15 k_{2}=63\end{array}\right.$
to get $\left\{\begin{array}{l}k_{1}=3, \\ k_{2}=-4\end{array}\right.$
$$
\therefore y=\frac{3}{x}+4(x-2) \text {. }
$$

Substitute $x=-5$ into the above equation, we get $y=-28 \frac{3}{5}$.
Answer: When $x=-5$, the corresponding value of $y$ is $-28 \frac{3}{5}$.