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.
 A: Guess and check is a valid method of solving equations. Notice that the domain of square root necessitates that
$$x-y \geq 9$$
so what happens if $x-y = 9$ ?
$$\begin{cases}3-(y+1)^2 = 3 \\ x+8y = 0\end{cases}$$
Can you take it from here?
Notice that the first equation also mandates that $-1-\sqrt{3} \leq y \leq -1+\sqrt{3}$ because the outcome of a square root always needs to be positive. Can you use this to prove there can't be any other solutions?
A: Since $x-y\geq9,$ we obtain:
$$3-(y+1)^2=\sqrt{x-y}\geq3,$$ which gives $$y=-1,$$
$$x-y=9$$ and $$x=8.$$
Now, it's enough to check that the second equation holds for these values of $x$ and $y$.
A: 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.
