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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.

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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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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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  • $\begingroup$ Your solution is really bad, i do not want to rely too much on calculation support software. $\endgroup$ – nDLynk Mar 8 at 15:48
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    $\begingroup$ Ok i'm searching for a new one, one moment please. $\endgroup$ – Dr. Sonnhard Graubner Mar 8 at 15:49
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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 $(1)$

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