# How to solve an overdetermined system of equations defining the dependent variables?

I'm trying to solve a simple system of six equations and four variables, namely, $F,G,K$ and $L$. I need to solve the system for $F$ and $L$ in terms of $G$ and $K$,that is

$$F(G,K) \,\, ,\\ L(G,K) \,\, ,$$ but I don't have any idea of how can I command the Mathematica do it. Lately I'm using the command Solve to do that.

This is the system of equation that I'm treating with:

eq1=F + G == H + J

eq2=I F ω - I G ω == I H Sqrt[ω^2 - Subscript[V, 0]] - I J Sqrt[ω^2 - Subscript[V, 0]]

eq3=E^(I a Sqrt[ω^2 - Subscript[V, 0]]) H + E^(-I a Sqrt[ω^2 - Subscript[V, 0]]) J ==  E^(I a ω) K + E^(-I a ω) L

eq4=I E^(I a Sqrt[ω^2 - Subscript[V, 0]]) H Sqrt[ω^2 - Subscript[V, 0]] - I E^(-I a Sqrt[ω^2 - Subscript[V, 0]]) J Sqrt[ω^2 - Subscript[V, 0]] == I E^(I a ω) K ω - I E^(-I a ω) L ω

eq5=E^(I b ω) K + E^(-I b ω) L == 0

eq6=I E^(I b ω) K ω - I E^(-I b ω) L ω == 0


• In fact you want to find the solution of 6 equations with respect to 4 variables: F, L, ,J and H. The fact that you are not interested in the values of J and H plays no role. In this case, however, your system is over-determined. You only need 4 equations to do that. Then Solve will do the job. Note that K is reserved in Mma, you should not use it. – Alexei Boulbitch Aug 5 '16 at 8:50
• @AlexeiBoulbitch, Thank you mate, I think that my problem now is with mathematics. Can you draft and solution for this problem? – Herr Schrödinger Aug 5 '16 at 8:58
• No. As I explained above the mathematical statement of the problem is incorrect. I do not know, how to make a correct statement, since it depends on the problem you solve. You know this problem, while I do not. – Alexei Boulbitch Aug 5 '16 at 9:07
• I'll edit the question to fix the mathematical statement. – Herr Schrödinger Aug 5 '16 at 9:09

I redefined the equations with lower cases.

eq = {eq1, eq2, eq3, eq4, eq5, eq6};

Solve[Eliminate[eq, l], f]


{{f -> -g + h + j}}

Solve[Eliminate[eq, f], l]


{{l -> -E^(2 I b \ω) k}}

• It didn't work, I need $F(G,K)$ and $L(G,K)$..... – Herr Schrödinger Nov 3 '16 at 14:26