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I have three different equations of the form below, with two variables, x and y, and all other letters are constants. I want an algorithm that solves for x and y under any real values of the constants. Thanks for any help!

$a_1x^3 + b_1x^2y + c_1xy^2 + d_1y^3 + e_1x^2 + f_1x + g_1y^2 + h_1y + k_1 = 0$

$a_2x^3 + b_2x^2y + c_2xy^2 + d_2y^3 + e_2x^2 + f_2x + g_2y^2 + h_2y + k_2 = 0$

$a_3x^3 + b_3x^2y + c_3xy^2 + d_3y^3 + e_3x^2 + f_3x + g_3y^2 + h_3y + k_3 = 0$

I'm hesitant to substitute to reduce from cubics to quadratics, because I don't know how to avoid missing a root in the process.

I found a link to a book by Sturmfels, but I have only high school math, and I was not able to identify an answer here: https://math.berkeley.edu/~bernd/cbms.pdf

I found this related discussion, but its algebraic manipulation between polar and cartesian expressions is more than I could create to solve something like this: Solving a system of two cubic equations

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  • $\begingroup$ Any software able to solve systems of equations would be able to solve this, for instance Mathematica (and probably Maple as well). $\endgroup$ – Bobson Dugnutt Apr 17 '17 at 20:54
  • $\begingroup$ Are you speaking about real roots because the probability that a system of 3 cubic equations have a common real root is 0. You are in high school, but have you already studied complex numbers ? $\endgroup$ – Jean Marie Apr 17 '17 at 20:54
  • $\begingroup$ each of the three equations should have an $xy$term and a constant. As written, the origin lies on all three $\endgroup$ – Will Jagy Apr 17 '17 at 21:31
  • $\begingroup$ to continue JeanMarie's point, you should solve just two such equations. $\endgroup$ – Will Jagy Apr 17 '17 at 21:35
  • $\begingroup$ The idea is to write some code, first in MS Excel and then in PHP, for processing answers from a lot of data that generate these few constants. So I'm hoping for an algorithm to process many, many sets. This is part of a bigger problem to find an algorithm for minimums in analyzing certain data sets. I've gotten to this point -- the equation forms above -- with partial derivatives and substitution to eliminate a third variable. (I'm a 47yo math teacher, career-switcher who never went past calculus.) $\endgroup$ – ptme Apr 17 '17 at 21:55

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