Question: > 
> 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.

Answer: 
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?
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 $y$.