<s>[INST] Solving the system with $x^2 -xy -1=0$ and $2xy-4y^2 +3=0$ I want to solve the following system of quadratic equations: 
$$\begin{align}
x^2 -xy -1 &=0 \\ 2xy-4y^2 +3 &=0
\end{align}$$
How can I do it?
My attempt:
$$xy = 1 -x^2$$
$$2x^2=4y^2-1$$
$$x = \sqrt{2y^2 - 0.5}$$
After plugging $x$, I get an equation that I can't solve.
 [/INST] $$xy=x^2-1$$
$$2(x^2-1)=4y^2-3$$
$$2x^2=4y^2-1$$
$$4y^2=2x^2+1\tag{3}$$
From the first equation, 
$$x^2-1=xy$$
Let's square both sides and use $(3)$,
$$(x^2-1)^2=x^2y^2=x^2\left(\frac12 x^2 + \frac14\right)$$
$$x^4-2x^2+1=\frac12 x^4 + \frac14x^2$$
$$\frac12 x^4-\frac94x^2+1=0$$
$$2x^4-9x^2+4=0$$
$$(2x^2-1)(x^2-4)=0$$
I will leave the rest to you as an exercise.
</s>