<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] 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 $y=8$.
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