[Algebra]
Problem: If $x$ and $y$ are real numbers, determine all solutions $(x, y)$ of the system of equations

$$
\begin{aligned}
& x^{2}-x y+8=0 \\
& x^{2}-8 x+y=0
\end{aligned}
$$

Solution: Subtracting,

$$
\begin{array}{r}
x^{2}-x y+8=0 \\
x^{2}-8 x+y=0 \\
\hline-x y+8 x+8-y=0 \\
8(1+x)-y(1+x)=0 \\
(8-y)(1+x)=0 \\
y=8 \text { or } x=-1
\end{array}
$$



If $y=8$, both equations become $x^{2}-8 x+8=0, x=4 \pm 2 \sqrt{2}$.

If $x=-1$ both equations become $y+9=0, y=-9$.

The solutions are $(-1,-9),(4+2 \sqrt{2}, 8)$ and $(4-2 \sqrt{2}, 8)$.
If $x^{2}-x y+8=0, y=\frac{x^{2}+8}{x}$.

And $x^{2}-8 x+y=0$ implies $y=8 x-x^{2}$.

Equating, $\frac{x^{2}+8}{x}=8 x-x^{2}$

$$
\text { or, } x^{3}-7 x^{2}+8=0 \text {. }
$$

By inspection, $x=-1$ is a root.

By division, $x^{3}-7 x^{2}+8=(x+1)\left(x^{2}-8 x+8\right)$.

As before, the solutions are $(-1,-9),(4 \pm 2 \sqrt{2}, 8)$.

Final answer: $(-1,-9),(4+2 \sqrt{2}, 8),(4-2 \sqrt{2}, 8)$