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?