<s>[INST] Solving the system $3-(y+1)^2 = \sqrt{x-y}$, $\;x+8y = \sqrt{x-y-9}$ 
How to solve these equations?
$$\begin{cases}
3-(y+1)^2 = \sqrt{x-y}\\
x+8y = \sqrt{x-y-9}
\end{cases}$$

I've tried solving this using the substitution and elimination methods without any success. I also tried plotting these equations and I got $x = 8$, $y = -1$.
Can someone show me the steps required to solve this?
Thanks in advance.
 [/INST] Observing the right-hand sides, we get:
$$(x+8y)^2 + 9 = (3 - (y+1)^2 )^2$$
$$(x+8y)^2 -(3 - (y+1)^2 )^2 = -9$$
and now using the difference of two squares:
$$(x+8y+3-(y+1)^2)(x+8y-3 + (y+1)^2 )= -9$$
If there is a clean solution where $x, y$ are integers, then the two brackets must be integers themselves. There are only a few possibilities: which are: $$(\text{left}, \text{right}) = (-1, 9), (1, -9), (-3, 3), (3, -3), (-9, 1), (9, -1).$$
Some of these solutions are extraneous or have non-integer solutions (that can be expressed in radicals). With the pair $(3, -3)$, you get an integer solution $(x,y) = (1,8)$, and now substitute into the original equations to verify them.
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