<s>[INST] A 3-minute algebra problem I have just taken advance math test level 2 and there are several problems that have been bugging me. This is the first question:

If $x,y>0$, then determine the value of $x$ that satisfies the system
  of equations: \begin{align} x^2+y^2-xy&=3\\ x^2-y^2+\sqrt{6}y&=3\\
\end{align}

I can answer this problem using a 'standard' algebra but it takes time more than 3 minutes. For the sake of curiosity, is there a way to answer this problem less than 3 minutes?
 [/INST] 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 $x=\sqrt{2}$.
</s>