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.

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 $(1)$
