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Sep 22, 2015 at 4:28 answer added Eric Towers timeline score: 2
Sep 22, 2015 at 3:45 history tweeted twitter.com/#!/StackMma/status/646168359273410560
Sep 22, 2015 at 0:51 comment added Dr. belisarius Simpler Solve[{x == 1, dbtot == Sqrt[x^2], dib == dbtot}, {dib}]
Sep 21, 2015 at 23:51 comment added Bob Hanlon Use ToRules with Reduce to get rules rather than equations as results.
Sep 21, 2015 at 23:41 comment added David G. Stork @JackLaVigne Thanks... that is a help, but still a bit unsatisfactory. After all, I would like to extract the value of dib using dib /. Reduce[...] but that will not work elegantly.
Sep 21, 2015 at 23:41 answer added Bob Hanlon timeline score: 5
Sep 21, 2015 at 23:33 comment added Jack LaVigne Replacing Solve with Reduce handles both your cases. I haven't the foggiest as to why Solve is struggling.
Sep 21, 2015 at 22:55 history edited David G. Stork CC BY-SA 3.0
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Sep 21, 2015 at 22:42 history edited David G. Stork CC BY-SA 3.0
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Sep 21, 2015 at 22:39 comment added David G. Stork @MarcoB Thanks, but I'd really like to understand why the straightforward approach simply does not work for a set of very elementary equations and variable assignments. There should be no need to be smart about eliminating variables in such a situation.
Sep 21, 2015 at 22:33 comment added MarcoB It is weird at first blush. I think I am missing something here. Anyway, you can ask Solve to explicitly eliminate dbtot from the system of equations, and then it will return your expected result: Solve[eq, dib, {dbtot}], where eq is your system of equations, returns {{dib -> -0.178513}}.
Sep 21, 2015 at 22:25 history asked David G. Stork CC BY-SA 3.0