# 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.

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 $$y=8$$.

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.

• Your solution is really bad, i do not want to rely too much on calculation support software. – nDLynk Mar 8 '19 at 15:48
• Ok i'm searching for a new one, one moment please. – Dr. Sonnhard Graubner Mar 8 '19 at 15:49

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 $$(1)$$