Solving the system with $x^2 -xy -1=0$ and $2xy-4y^2 +3=0$

Question

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.

Answer 1

$$xy=x^2-1$$ $$2(x^2-1)=4y^2-3$$

$$2x^2=4y^2-1$$

$$4y^2=2x^2+1\tag{3}$$

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.

Answer 2

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

$$(\pm\frac1{\sqrt2},\mp \frac1{\sqrt2}),\>\>\> (\pm2,\pm \frac3{2})$$