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In Wolfram Mathematica 12, when I enter:

Solve[ x == y , x]

I get:

{{x -> y}}

as expected.

But when I add an equation that contains only constants and parameters:

Solve[ x == y z && z == 1, x]

I get no results:

{}

when I want to get {{x -> y}} as before.

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    $\begingroup$ I think this is known and intentional. Try Solve[x == y z && z == 1, {x, z}] I think I saw question on this before on this site but can't be sure now. $\endgroup$ Commented Apr 27, 2020 at 10:26
  • $\begingroup$ Thanks a lot, Solve[x == y z && z == 1, {x, z}] gives {{x -> y, z -> 1}}. This is (arguably more important) half of the answer. I would like to know why Mathematica behaves in this way. $\endgroup$ Commented Apr 27, 2020 at 10:34
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    $\begingroup$ You should evaluate Solve[ x == y z && z == 1, x,MaxExtraConditions->All]. Your problem is a special case of a general issue which can be found in What is the difference between Reduce and Solve? $\endgroup$ Commented Apr 27, 2020 at 10:38
  • $\begingroup$ see help, under possible issues. screen shot !Mathematica graphics Solve[x == 1 && a == 2, x] gives {} $\endgroup$ Commented Apr 27, 2020 at 10:39
  • $\begingroup$ You can also use either Solve[x == y z && z == 1, x, {z}] or Solve[Eliminate[x == y z && z == 1, z], x] $\endgroup$ Commented Apr 27, 2020 at 14:38

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