Solving System of Equations Symbolically

I have a more complex system of equations than the one below, but this one has the same issue.

I want to solve for a variable in terms of other variables.

Why does this not work to solve for c?

clc
syms a b c d e
eqn1= a==b+d;
eqn2= a==e+c;
eqnarray=[eqn1 eqn2];
solve(eqnarray,c)

Clearly the answer is c=a-e -> c=b+d-e

I'm unsure as to why I'm having so much trouble with this, I've solved systems of equations before symbolically, though I've never either had free variables or overdetermined systems.

I probably could have solved my full system of equations by hand already, so I'm feeling a little slow today.

Answer 1

[Star Strider]

Subtract them:

syms a b c d e
eqn1= a==b+d;
eqn2= a==e+c;
eqnarray=[eqn1 - eqn2];
cs = solve(eqnarray,c)

producing:

cs =
b + d - e

[Michal Walko]

Okay, that may work for this case, but what if my system of equations was not defined purely in terms of a single variable on the LHS such as below?

syms a b c d e
eqn1= b==a-d;
eqn2= a==e+c;
eqnarray=[eqn1 eqn2];
cs = solve(eqnarray,c)

EDIT: the subtraction thing works in this case again- why is this? I've never seen anything of this format in the documentation

Thanks!

[Star Strider]

My approach works in this situation as well:

syms a b c d e
eqn1= b==a-d;
eqn2= a==e+c;
eqnarray=[eqn1 - eqn2];
cs = solve(eqnarray,c)

producing:

cs =
2*a - b - d - e

EDIT —

Subtracting them works by creating one algebraic equation from the pair of algebraic equations. So long as they have at least one variable in common (as they do here), you can solve for the variable of interest.

[Michal Walko]

Edited my post before I saw your reply.

Thank you for your help, in some situations MATLAB accepts the system of equations purely as an array of form

equations=[first second third fourth] 

and gives me a solution if I want to solve for a variable inside.

Why does that not work in this case and do you know if this subtraction method has a name in the MATLAB documentation?