Solving a simultaneous equation How can I solve the following simultaneous equations:
$$3x^4+3x^2y^2-6xy = 0\tag 1$$
$$-2x^3y+3x^2-y^2=0\tag 2$$
I have tried rearranging for $y$ in eq(1) and plugging it into eq(2), but the result did not give me the right answer.
 A: Apart the obvious solution $(x=0,y=0)$ :
$$x^3+xy^2-2y=0$$
$$3x^2-y^2-2x^3y=0$$
From eq.(2) : $y^2=3x^2-2x^3y$ that is plugg into eq.(1) 
$x^3+x(3x^2-2x^3y)-2y=0$
$$y=\frac{2x^3}{x^4+1}$$
Plugging it into eq.(1) leads to :
$$x^8+2x^4-3=0$$
$$x^4=-1\pm 2$$
$$x=(-1\pm 2)^{1/4}$$
$$y=\frac{2x^3}{x^4+1}=(-1\pm 2)^{3/4}$$
So, eight solutions (real and complex).
The real solutions are $(x=1,y=1)$ and $(x=-1,y=-1)$ 
A: HINT: eliminating $y$ we get for $x$:
$$x \left( x-1 \right)  \left( x+1 \right)  \left( {x}^{2}+1 \right) 
 \left( {x}^{4}+3 \right) 
=0$$
A: Using the old Sylvester´s method of elimination, choosing $y$ (the lowest degree) to be eliminated, we have
$$\begin{vmatrix}
3x^2&-6x&3x^4&0\\
0&3x^2&-6x&3x^4\\
1&2x^3&-3x^2&0\\
0&1&2x^3&-3x^2
\end{vmatrix}=0$$ which gives $$36x^4(x^8+2x^4-3)=0$$ i.e.
$$x^4(x^4+3)(x^2+1(x+1)(x-1)=0$$
The $12$ values for $x$ are $$x=0\space \text{of order four}$$   $$x=\pm 1$$ $$x=\pm i$$ $$x= \pm \sqrt[4]{-3}$$ $$x=\pm i\sqrt[4]{-3}$$
(where $\sqrt[4]{-3}=\frac{(1+i)\sqrt[4]{-3}}{\sqrt 2}$)
To each of these values of $x$ it correspond correlative values of $y$.
