Suppose there is an equation like this (overly simplified example):

eqn1 = ui - ua r1 / (r1 + r2) == 0

How can I get Mathematica to "solve" the equation in terms of:

eqn2 = v == ua / ui

i.e. have Mathematica deliver v == 1 + r2 / r1 as a result? Despite googling I cannot find the proper way (tried equation system, Eliminate, LinearSolve, ...). Thanks in advance - Rob


You can combine Solve with Eliminate:

    Eliminate[{ui-ua r1/(r1+r2)==0,v==ua/ui}, ua],

{{v -> (r1 + r2)/r1}}

  • $\begingroup$ works like a charm, thx. Shouldn't the 2nd parameter on Eliminate be the list {ua , ui} rather than {ua} ? $\endgroup$ – Robert GID S. May 30 at 0:03
  • $\begingroup$ @RobertGIDS. With only two equations, I think it's safer to try to eliminate just one variable. $\endgroup$ – Carl Woll May 30 at 0:17

Although not currently documented, you can include in Solve a list of variables to be eliminated.

Solve[{ui - ua r1/(r1 + r2) == 0, v == ua/ui}, v, {ua}][[1]] // Expand

(* {v -> 1 + r2/r1} *)
  • $\begingroup$ Thx, this works indeed - though I'm confused as the Language Reference designates Solve's 3rd parameter as "domain", e.g. Integers, Reals, Complexes - it never occured to me that "domain" could also be used for variable elimination. I guess not many folks are aware of this context - so kudos to you for pointing this out. I'd appreciate if you kindly comment on this "[ [1] ]" construct - what is it's meaning ? $\endgroup$ – Robert GID S. May 29 at 23:56
  • $\begingroup$ @RobertGIDS. - As I stated, the use of a list of variable to be eliminated is not documented. It was a long time ago and presumably still functions to maintain backward compatibility. Consequently, the third argument here is not being used to specify a domain. The [[1]] takes the first element of a list (see Part). In this case it removes an extra set of list brackets. $\endgroup$ – Bob Hanlon May 30 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.