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$f(x, y) = 0$ and $g(x, y) = 0$, both $f$ and $g$ are cubic polynomial equation (at most 10 coefficients for each).

Is there any fixed method to solve this degenerate equation system? thanks.

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closed as unclear what you're asking by Andrey Rekalo, David White, Daniel Moskovich, John Pardon, Chris Godsil Jul 30 '13 at 18:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

WHat's degenerate about it? And what do you mean by "solve"? Numerically? In radicals? – Igor Rivin Jul 30 '13 at 13:43

Resultants are good, but for practical computations the Bezout Lemma (it's the same as the extended Euclidean algorithm) can be less expensive: Assume that $f(x,y)$ and $g(x,y)$ are relatively prime. Consider $f$ and $g$ as polynomials in $y$ over the field $k(x)$, where $k$ is your (unspecified) base field. Then there are $r,s\in k(x)[y]$ with $r(y)f(x,y)+s(y)g(x,y)=1$. Multiplying with the least common multiple $u(x)$ of the denominators of the coefficients of $r(y)$ and $s(y)$ yields $R(x,y)f(x,y)+S(x,y)g(x,y)=u(x)$, where $R,S\in k[x,y]$. So if $f=g=0$, then $u=0$.

Of course $u(x)$ is essentially the resultant of $f$ and $g$ with respect to $y$.

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I'd suggest reading about resultants. The applications section of that article gives a method for solving your problem.

P.S. The level of this question seems maybe borderline for this forum (non-research-level questions can always be asked at, but I could imagine someone going through grad school without learning about resultants so it seemed worth answering.

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The magic word is "resultant" (which see, on, e.g., Wikipedia). The resultant of your equations will be a single one variable equation of degree nine. There is not much hope of it being solvable in radicals, but it will be easily solved numerically. On the other hand, if you have software which will solve it numerically, it will probably solve the original system (in Mathematica I believe it is "NSolve")

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