Solve a system of polynomial equations

Question

I cannot find a way to solve the following polynomial equation system

$$\begin{cases}-3x^{2}+y^{2}-2xy-\alpha_{1}x+\alpha_{0}=0\\ -x^{2}+3y^{2}+2xy-\beta_{1}y+\beta_{0}=0\end{cases}$$

Could you help me to solve this?

Answer 1

If you have two equations in two variables, a neater way is to eliminate one variable using resultants. (Click on this WA link.) Eliminating $y$, one gets (rather suprisingly) a cubic in $x$,

$$ (32 a_1+32 b_1)x^3+ (10 a_1 b_1+9 a_1^2-32 a_0-3 b_1^2+32 b_0)x^2+ (-a_1 b_1^2-10 a_0 b_1+6 a_1 b_0-18 a_0 a_1-2 b_0 b_1)x+(a_0 b_1^2-6 a_0 b_0+9 a_0^2+b_0^2)=0$$

Thus, for any real $a_0,\,a_1,\,b_0,\,b_1$ you use, at least one of $x,y$ will also be real. The cubic can be put in the form,

$$ax^3+bx^2+cx+d=0$$

and a solution for $x$ is easily found. Unfortunately, you have four parameters, so the end symbolic result will still be messy, though it will not be too bad for numerical examples.

$\color{green}{(Edit)}$

Example: Let $\alpha_1,\alpha_0 = 4,\,3$ and $\beta_1,\beta_0 = 6,\,1$. So,

$$-3x^2 + y^2 - 2x y - 4x + 3 = 0\\-x^2 + 3y^2 + 2x y - 6y + 1 = 0$$

Eliminate $x$ using WA and we get,

$$4 - 8 y - 113 y^2 + 80 y^3=0\tag1$$

and eliminating $y$,

$$43 - 132 x + 53 x^2 + 80 x^3=0\tag2$$

However, $(1)$ and $(2)$ will have three solutions each, so one still has to find the correct pairs $x_i,y_i$ by numerical testing. Both are 3-real root cubics, and the correct pairs are,

$$x_1,\, y_1 \approx -1.767833,\; -0.209114\\ x_2,\, y_2 \approx 0.515306,\; 0.1640421\\ x_3,\, y_3 \approx 0.590026,\; 1.457572$$

Of course, one can then find the exact expressions for the $x_i,y_i$ by using the cubic formula.