The lines with the two equations below intersect at the point $(2,-3)$.

$$
\begin{array}{r}
\left(a^{2}+1\right) x-2 b y=4 \\
(1-a) x+b y=9
\end{array}
$$

What are the possible ordered pairs $(a, b)$ ?
Substituting $x=2$ and $y=-3$ into the two equations gives

$$
\begin{aligned}
2\left(a^{2}+1\right)-2 b(-3) & =4 \\
2(1-a)+b(-3) & =9
\end{aligned}
$$

which can be simplified to get

$$
\begin{aligned}
2 a^{2}+6 b & =2 \\
2 a+3 b & =-7
\end{aligned}
$$

Dividing the first equation by 2 gives $a^{2}+3 b=1$ from which we can subtract $2 a+3 b=-7$ to get $a^{2}-2 a=8$.

Rearranging gives $a^{2}-2 a-8=0$ which can be factored to get $(a-4)(a+2)=0$.

Therefore, there are two possible values of $a$, which are $a=4$ and $a=-2$. Substituting these values into $2 a+3 b=-7$ and solving for $b$ gives $b=-5$ and $b=-1$, respectively.

The ordered pairs $(a, b)=(4,-5)$ and $(a, b)=(-2,-1)$ are the two possibilities. It can be checked that the two systems of equations

$$
\begin{array}{rlrl}
17 x+10 y & =4 & 5 x+2 y & =4 \\
-3 x-5 y & =9 & 3 x-y & =9
\end{array}
$$

both have the unique solution $(x, y)=(2,-3)$.

$$
\text { ANsWER: }(4,-5) \text { and }(-2,-1)
$$