# How to solve a system of two equations and two variables in function of others terms

I want to Find C1 and C2 in terms of a and e from the following system of equations:

60 + 31 a + 4 a^2 + 4 C1 C2 (4 + a) (873 + 444 a + 56 a^2) e + 128 C1^2 C2^2 (3 + a) (4 + a) (9 + 2 a) (33 + 8 a) e^2 - 3072 C1^3 C2^3 (4 + a)^2 (9 + 2 a)^3 e^3 == 0

60 + 31 a + 4 a^2 + C1 C2 (4 + a) (3267 + 8 a (207 + 26 a)) e + 16 C1^2 C2^2 (4 + a) (9 + 2 a) (837 + 468 a + 64 a^2) e^2 - 4 C1^3 C2^3 (4 + a) (9 + 2 a)^2 (18333 + 8472 a + 976 a^2) e^3 == 0


When I used "Solve",

Solve[{60 + 31 a + 4 a^2 + 4 C1 C2 (4 + a) (873 + 444 a + 56 a^2) e +
128 C1^2 C2^2 (3 + a) (4 + a) (9 + 2 a) (33 + 8 a) e^2 -
3072 C1^3 C2^3 (4 + a)^2 (9 + 2 a)^3 e^3 == 0, 60 + 31 a + 4 a^2 + C1 C2 (4 + a) (3267 + 8 a (207 + 26 a)) e + 16 C1^2 C2^2 (4 + a) (9 + 2 a) (837 + 468 a + 64 a^2) e^2 - 4 C1^3 C2^3 (4 + a) (9 + 2 a)^2 (18333 + 8472 a + 976 a^2) e^3 == 0}, {C1, C2}]


I got: {}

Thanks for helping!!!

• Use Reduce[eqns, {C1, C2}] – Bob Hanlon Jan 13 '19 at 18:56
• Each of the two equations is a cubic in C1*C2. In general, the two equations have no common roots, so Solve returns an empty set, { }. At least one of the two parameters a and e must assume a special value for the solution set not to be empty. See SolveAlways. – bbgodfrey Jan 13 '19 at 23:46
• Please do not use tags unrelated to the problem that are you asking about. – Szabolcs Jan 21 '19 at 15:56