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I have the following problem:

I have a system of equations specifically consisting of 56 equations, with 34 coefficients to be determined, in simplified form:

(a+2b-c)A[something]==0
(b+3d-c)A[something else]==0
...

Where A is some operator. Now, I want to use Solve to solve this system of equations:

Solve[(a+2b-c)A[something]==0&&(b+3d-c)A[something else]==0&&...,{a,b,c,d,....}]

However, this takes quite a long time because the expressions are somewhat complicated and lengthy.

I'm not exactly sure how the algorithm works and whether Solve[] only looks at the expressions in the parentheses or also works with the operators A[], which might lead to incorrect expressions because I defined the effects of A myself, so Mathematica might not know how to handle the A's properly.

In summary, my question is:

Does the Solve[] algorithm only interact with the expressions within parentheses, or does it do something else?

If not, is there a simple method to write something similar to Solve[] that leaves the operators A[] alone?

Any opinions and tips on speeding up the process are very welcome.

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    $\begingroup$ If you want to get the best answer and help, please include your whole code in the question. $\endgroup$
    – Domen
    Commented Apr 16 at 13:46
  • $\begingroup$ Compare the result from Solve with the result from Reduce. Solve is very good at some categories of problems. Reduce is better at others, but sometimes slower. Start with a small problem that is similar in form to your full problem, see if Reduce can crack it and study the result. Then make a slightly larger problem and repeat. When I don't get an instant solution from MMA for a problem I find it helpful to think "just imagine how much longer this would have taken if I were scratching this out on a piece of paper with a crayon" and I find myself MUCH happier at that point. $\endgroup$
    – Bill
    Commented Apr 16 at 15:29
  • $\begingroup$ Possibly quick experiment to see what happens if "Solve leaves the operators A[] alone" Restart MMA or clear any definitions of A[] and then try Solve or Reduce without defining A[anything]. That hopefully will leave all A[anything] alone and treated as a simple unknown and see if Solve or Reduce can give you a solution containing those unknowns. If that experiment doesn't work then try defining A[_]=1 before your Solve and that will show you a result as if the A[anything] just disappears. $\endgroup$
    – Bill
    Commented Apr 16 at 16:12

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