<s>[INST] How to calculate the nullity for a matrix with variable? In order to find the number of (x,y) satisfying  
$$f(x,y) =\begin{cases}
    xy^2-y+x^2+1=0\\
    x^2y^2+y-1=0\\
  \end{cases}$$
I use the Jacobi-Darboux Theorem by fix 1.x; 2.y and get the Bezoutian matrix, but I don't know how to find the nullity of the both matrix... 


*

*$$B = 
 \begin{pmatrix}
  y^3-y^2 & y^3-y^2+y-1\\
  y^3-y^2+y-1 & -y^4 
 \end{pmatrix}$$

*$H_{f_x} = 
 \begin{pmatrix}
  -1 & x\\
  x & 0 \end{pmatrix}$, $T_{f_x} = 
 \begin{pmatrix}
  1+x^2 & -1\\
  0 & 1+x^2 \end{pmatrix}$, $H_{g_x} = 
 \begin{pmatrix}
  1 & -1-x\\
  -1-x & 0 \end{pmatrix}$, $T_{g_x} = 
 \begin{pmatrix}
  -1 & 1\\
  0 & -1 \end{pmatrix}$
$$B = \begin{pmatrix}
  1 & -1-x\\
  -x & -x \end{pmatrix} - \begin{pmatrix}
  1+x^2 & -1+x^2+x^4\\
  x^2+x^4 & -x^2 \end{pmatrix}= 
 \begin{pmatrix}
  -x^2 & -x^4-x^2-x\\
  -x^4-x^2-x & x+x^2 
 \end{pmatrix}$$
 [/INST] We can solve the two polynomial equations using $S$-polynomials (from Buchberger's algorithm). Apart from the trivial solution $(x,y)=(0,1)$ this yields
$$
x=y^4 + y^3 - 2y^2 + y - 1,
$$
so that we obtain a polynomial in $y$, namely
$$
y^6 + y^5 - y^4 + 2y^3 - 2y^2 + y - 1=0.
$$
This has $6$ solutions over the complex numbers. Two of them are real solutions,
i.e., $y=0.863774039675$ and $y=- 2.15866388153$.
</s>