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I want to solve these equations but Mathematica gives no answer.

X := 
  3/2 x (-((γ y (1 - z)^3 (-x^2 - y^2 + 1)^2)/z^3) + x^2 - y^2 - 1) - 
    (λ y z)/(1 - z);

Y := 
  -((3 γ (y^2 - 1) (1 - z)^3 (-x^2 - y^2 + 1)^2)/(2 z^3)) + 3/2 y (x^2 - y^2 + 1) + 
    (λ x z)/(1 - z);

Z := 
  3/2 ((1 - z) z (x^2 - y^2 + 1) - (γ y (1 - z)^4 (-x^2 - y^2 + 1)^2)/z^2);

Solve[X == Y == Z == 0, {x, y, z}]

{}

What do I do?

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  • 3
    $\begingroup$ Reduce[{X == Y == Z == 0}, {x, y, z}] $\endgroup$
    – user36273
    Jun 25, 2016 at 10:25
  • 1
    $\begingroup$ As a side note, it is considered bad form to use capital letters as declared variables. Or to use multi-letter variables that start with capital letters. $\endgroup$
    – Feyre
    Jun 25, 2016 at 11:41
  • $\begingroup$ This system cannot be solved with the methods available to Solve. (or Reduce) $\endgroup$
    – Young
    Jun 26, 2016 at 3:03
  • $\begingroup$ @Young Do you Know another method? $\endgroup$
    – milad
    Jun 26, 2016 at 7:32
  • $\begingroup$ @milad , @Young This system can be solved with Reduce! $\endgroup$
    – user36273
    Jun 26, 2016 at 12:35

1 Answer 1

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Please check the documentation on Solve.

Under Details:

enter image description here

{} means that your system has no solutions in general.

Reduce does give a solution, but notice that it comes with the specific condition that λ == 0. For general λ there is no solution.

Solve gives generic solutions only. Another excerpt from the documentation:

Solve gives generic solutions only. Solutions that are valid only when continuous parameters satisfy equations are removed. Other solutions that are only conditionally valid are expressed as ConditionalExpression objects.

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