# Eliminating variables from system of equations

I have a system of equations for algebraic curve given by the zero locus of some polynomial encoded in the system of equations:

$$y = \frac{z^{1+a}}{z-x},$$ $$1= \frac{(x-z) x^a z^{-1-2a+b}}{z-1}$$

I want to eliminate variable $z$ to get equation for zero locus of some polynomial in $x$ and $y$. We can assume for constants $a,b$ to be integers (or even positive). I first tried to multiply first equation by the second but this doesn't really do much.

I would appreciate any good ideas.

• It seems like your equations lead to polynomials with a degree in terms of $a$ and $b$ - and since it's impossible to solve polynomials of degree higher than $4$, it seems unlikely (however not impossible) that your system has a nice solution.
– user304329
Commented Mar 12, 2017 at 21:26
• That's what I'm thinking, however I'm trying to find a nice way but so far I'm not lucky. Terms like ($(y+x)^{f(a,b)}$ would be OK too, but I can't even get those. Commented Mar 12, 2017 at 21:30

With $a,b$ non-negative integers, you can rewrite the two equations in polynomial form as:

\begin{align} y(z-x) - z^{a+1} & = 0 \\ z^{2a+1}(z-1) - (x-z) x^a z^{b} & = 0 \end{align}

Then eliminating $z$ between the two equations can be technically done using polynomial resultants but the calculations are rather heavy, and the results not too pretty. For example:

• $a=1, b=1\,$: $\;x^2 y^2 (x^4 + 2 x^3 y + x^2 y^2 - x^2 y - x y^2 + x y) = 0$

• $a=1, b=2\,$: $\;-x^3 y^3 (x^5 y - 2 x^3 y + x^2 y^2 + 3 x^2 y - x y^2 + x y - y + 1) = 0$

• $a=2, b=1\,$: $\;x^2 y^2 (-x^9 + 3 x^7 y + x^5 y^3 - 3 x^5 y^2 + x^4 y^4 - x^3 y^4 + x^3 y^3)=0$

• $a=2, b=2\,$: $\;x^4 y^4 (-x^9 - x^6 y + x^5 y + 3 x^4 y - 2 x^3 y + x^2 y^2 - x y^2 + x y)=0$

• $a=3, b=5\,$: $$-x^7 y^7 (x^{19} y^3 - 6 x^{13} y^3 + 7 x^{12} y^3 + x^{10} y^4 + 2 x^{10} y^2 - x^9 y^4 - 13 x^9 y^2 + 14 x^8 y^2 + 3 x^7 y^3 - 3 x^6 y^3 - 4 x^5 y + 3 x^4 y^2 + 7 x^4 y - 3 x^3 y^2 + x y - y + 1)=0$$

• But can we keep arbitrary a and b? Results should not be pretty I think Commented Mar 12, 2017 at 22:21
• You might be able to derive a closed form in $a,b$ since the original polynomials have few terms, so the Sylvester matrix whose determinant is the resultant will have many $0$ elements. I don't necessarily expect it to be easy or pretty, though.
– dxiv
Commented Mar 12, 2017 at 22:25
• So by using Sylvester matrix method I would treat my polynomials as polynomials in z and then calculate the determinant? Commented Mar 13, 2017 at 18:11
• @Caims Right, the coefficients (and therefore the determinant) would be polynomials in $x,y\,$.
– dxiv
Commented Mar 13, 2017 at 18:37