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There are 2 equations of 3 variables and I have simplified and eliminated the third varibale to ge the relationship of the first and the second variable. Codes are as this:

Remove["Global`*"]
equa1 = \[Mu] == 1.3333333333333334`*^-6 + 1.2`/H;
equa2 = H == B/\[Mu];
Solve[Eliminate[{equa1, equa2}, {H}], \[Mu]]

The hint is

Solve was unable to solve the system with inexact coefficients. The
answer was obtained by solving a corresponding exact system and
numericizing the result.

But when I change to

Solve[Eliminate[{equa1, equa2}, {H}], B]

Then it is normal, so why?

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  • $\begingroup$ Try: equa1 = \[Mu] == Rationalize[1.3333333333333334*^-6 + 1.2/H, 0]; Solve likes exact numbers. $\endgroup$ Nov 19, 2022 at 11:24
  • $\begingroup$ This is no error, only a warning. MMA rationalized your machine numbers for the calculation. $\endgroup$ Nov 19, 2022 at 11:26

1 Answer 1

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In this example,we need not use Eliminate,using {μ}, {H} in Solve to solve {μ} and at the same time eliminate {H}.

Clear["Global`*"];
equa1 = μ == 1.3333333333333334`*^-6 + 1.2`/H;
equa2 = H == B/μ;
Solve[Rationalize[{equa1, equa2}, 0], {μ}, {H}]

enter image description here

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