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.

The given equations 
$$\begin{align}
& x^2 -xy = 1\tag1\\ &2xy-4y^2 = -3
\end{align}$$
lead to 
$$3(x^2 -xy)+ (2xy-4y^2)=0\implies (x+y)(3x-4y)=0$$
Then, substitute  $x+y=0$ and $3x-4y=0 $ into (1) to obtain the
solutions 
$$(\pm\frac1{\sqrt2},\mp \frac1{\sqrt2}),\>\>\> 
(\pm2,\pm \frac3{2})$$
