<s>[INST] 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.
 [/INST] 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 $(x=-1,y=-1)$ 
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