I would like to mechanically Solve[a y == (1 - a) x && u0 == x^a y^(1 - a),{x,y}], given 1>a>0 and u0>0. The emphasis is on mechanically: I know it is simple in this case to do a log transform and get solution to the transformed system. I'd like a simple approach to the original system. I want only the positive real solution.

But Solve encounters a problem:

Solve::incnst: "Inconsistent or redundant transcendental equation. After reduction, the bad equation is x - a x - a y == 0."

Restricting the domain to Reals does not help. As well as learning the proper approach to this problem, I would appreciate better understanding why the system is seen as inconsistent or redundant.

  • $\begingroup$ That is a weird ( and really incorrect ) error message. The problem is that the system can only be solved for a in the specified range, but Solve has no mechanism to specify that restriction. $\endgroup$
    – george2079
    Jul 20, 2015 at 19:33
  • $\begingroup$ @belisarius Why can you make the assumption that $x$ and $y$ are non-zero and positive? Also, is it as a consequence of that assumption that you canceled out an $x$ factor in the first equation? $\endgroup$
    – MarcoB
    Jul 21, 2015 at 1:21
  • $\begingroup$ @belisarius Oh I see; no, I think you are right in that interpretation. I missed that in the OP. Nevertheless, when I try to use Reduce on the OP's expression (with the $x$ in the first equation), and adding your conditions, MMA churns and churns but doesn't really return a solution. Does it work on your side? $\endgroup$
    – MarcoB
    Jul 21, 2015 at 1:37
  • $\begingroup$ @MarcoB deleting comments. I pasted the equation wrongly $\endgroup$ Jul 21, 2015 at 14:01

1 Answer 1


As you're interested only in Real and positive solutions, the following will do:

eq = {a y == (1 - a) x && u0 == x^a y^(1 - a)};
assum = {y > 0 && x > 0 && 1 > a > 0};
s0 = Solve[eq[[1, 1]], x];
sy = Solve[FullSimplify[eq[[1, 2]] /. s0[[1]], Assumptions -> assum], y];
js = Join[s0 /. sy, sy] // Flatten

(* {x -> -(((-1 + a^-1)^a a u0)/(-1 + a)), 
    y ->    (-1 + a^-1)^a u0}
  • $\begingroup$ @Alan Check it now, please $\endgroup$ Jul 21, 2015 at 13:59
  • $\begingroup$ Although this is not quite as "mechanical" as I hoped, it gets me there without explicit transformation of the functional form. The key was the FullSimplify after the substitution. Thanks. I recommend editing so that eq and assum are not lists, so you won't need multiple indexes for Part. Thanks. $\endgroup$
    – Alan
    Jul 21, 2015 at 15:31
  • $\begingroup$ @Alan Thanks! As you know, in this site you can edit other people's posts if you find it convenient! $\endgroup$ Jul 21, 2015 at 16:02

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