<s>[INST] 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.
 [/INST] 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. 
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