A 3-minute algebra problem

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Substracting second equation to the first you obtain $$2y^2-xy-\sqrt{6}y=0$$ Then, factoring $$y(2y-x-\sqrt{6})=0$$ Thus $2y-x-\sqrt{6}=0$ cause $x,y$ are greater than $0$. So $y=\frac{x+\sqrt{6}}{2}$, putting this expression into the second equation you have \begin{align} x^2-\left(\frac{x+\sqrt{6}}{2}\right)^2+\sqrt{6}\left(\frac{x+\sqrt{6}}{2}\right)&=3 \\ x^2-\frac{x^2+2\sqrt{6}x+6}{4}+\frac{\sqrt{6}}{2}x+3&=3\\ \frac{3}{4}x^2-\frac{3}{2}&=0\\ \end{align} Whose positive root is $x=\sqrt{2}$.

Take G.Bach's hint so that $2y=x+\sqrt 6$

Now add the two equations to obtain $$2x^2+(\sqrt 6-x)y=6$$ Substitute for $y$: $$2x^2+\frac 12(\sqrt 6-x)(\sqrt 6+x)=6=\frac 32 x^2+3$$

So that $x^2=2$