Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.
 A: From the second equation we get
$$\sqrt{x+y}=x+3-\sqrt{x}$$ y squaring this equation we obtain
$$y=(x+3-\sqrt{x})^2-x$$ plugging this in the first equation we get
$$\sqrt{(x+3-\sqrt{x})^2}+\sqrt{x+3}=\frac{1}{x}((x+3-\sqrt{x})^2-x-3)$$
Can you proceed?
Hint: $$x=1,y=8$$
Ok, we can also eliminate the square root:
$$\sqrt{x+y}=x+3-\sqrt{x}$$ so
$$x+3-\sqrt{x}+\sqrt{x+3}=\frac{1}{x}(y-3)$$
this is only a littlebit better. 
A: Let $\sqrt{x+y}=a\ge0,\sqrt{x+3}=b\ge0$
$$(b^2-3)(a+b)=a^2-b^2$$
If $a+b=0,a=b=0$
Else $b^2-3=a-b\iff a=b^2+b-3=x+b\iff x=a-b\  \  \  \ (1)$
Squaring we get
$$x+y=x^2+x+3+2bx$$
$$\iff y=x^2+3+2x\sqrt{x+3}$$
The value of $x,y$ must satisfy $(1)$
A: Remember that $$ \sqrt{a}+\sqrt{b}= {a-b\over \sqrt{a}-\sqrt{b}}$$ 
so $$\sqrt{x+y}+\sqrt{x+3}={y-3\over \sqrt{x+y}-\sqrt{x+3}} $$
and now we have $${y-3\over \sqrt{x+y}-\sqrt{x+3}}=\frac{1}{x}\left(y-3\right)$$
Obviuosly $y\neq 3$ so we have $$\sqrt{x+y}-\sqrt{x+3}=x$$
Since  $$\sqrt{x+y}+\sqrt{x}=x+3$$
we have now $$\underbrace{\sqrt{x}+\sqrt{x+3}}_{f(x)} = 3\implies x= 1$$
($f$ is strictly increasing function so this equation has at most $1$ solution.) And then we get $y=8$.
