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Solve the system of equations: $\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}+\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$


I used wolframalpha.com and got the only solution: $(x;y)=(4;4)$

And I guess that we can get $x=y$ from first equation.

And this is my try

We have $\sqrt{x^2+(y-2)(x-y)}-y+\sqrt{xy}-y=0$

$\Leftrightarrow (x-y)\left(\dfrac{x+2y-2}{\sqrt{x^2+(y-2)(x-y)}+y}+\dfrac{y}{\sqrt{xy}+y}\right)=0$

But I can't prove that $\dfrac{x+2y-2}{\sqrt{x^2+(y-2)(x-y)}+y}+\dfrac{y}{\sqrt{xy}+y}\ne0$.

So who can help me?

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  • $\begingroup$ from first equation, you can't get $x=y$, because $x=2,y=0$ is hold first equation $\endgroup$ – math110 Jun 23 '15 at 9:59
  • $\begingroup$ So how to solve this system of equations? $\endgroup$ – idiots Jun 23 '15 at 10:06
  • $\begingroup$ $(x=2,y=0)$,$(x=y)$ and $(x=0 , y=2/5)$ are 3 of the solution I can see for the 1st equation. $\endgroup$ – PleaseHelp Jun 23 '15 at 10:19
  • $\begingroup$ And @math110 $(x=2,y=0)$ and $(x=y)$ can both satisfy the equation, it is not a function. $\endgroup$ – PleaseHelp Jun 23 '15 at 10:22
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If we allow the non-principal square root, you can actually have more than one solution to the system, $$\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}\color{red}\pm\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$$ The positive case is solved by $x=y = 4$, but the negative case can be solved by,

$$x=2.771913334036360207047659037932620797468\dots\\ y=0.4518271741040497609916929241427121641268\dots$$

which are roots of $12$-deg equations and can be found by using resultants.

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