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I would like to have solutions of inequalities where condition of existence are included, but Reduce doesn't work like this, it would simplify the denominators before solving, like this:

Reduce[(x^2 + x)/x >= 0, Reals]
(* x >= -1 *)

Is there a way to have the solution with condition of existence of the original expression?

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  • $\begingroup$ Reduce[Exists[x, (x^2 + x)/x >= 0], Reals] gives True. Is this what you need? $\endgroup$ – Roman Mar 3 at 14:56
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I think this is what you mean:

Simplify[
  Reduce[(x^2 + x)/x >= 0, x] && FunctionDomain[(x^2 + x)/x, x]
]

(* Out: x >= -1 && x != 0 *)
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  • $\begingroup$ Doesn't always work, for example with "FunctionDomain[x/x, x] " I get the output "True" $\endgroup$ – Marco Disce Mar 4 at 13:53
  • $\begingroup$ @MarcoDisce That's because x/x will immediately evaluate to 1. You could use FunctionDomain[Hold[x/x], x] instead, which prevents evaluation. $\endgroup$ – MarcoB Mar 4 at 14:02

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