Solving a simultaneous equation

Question

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.

Answer 1

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 $(x=-1,y=-1)$

Answer 2

HINT: eliminating $y$ we get for $x$: $$x \left( x-1 \right) \left( x+1 \right) \left( {x}^{2}+1 \right) \left( {x}^{4}+3 \right) =0$$

Answer 3

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 $y$.