i'm new in the world of mathematica. i have a simple problem : i want to find a solution of a system of equations. The problems are:

1) Is it better to use Solve or Reduce (for this kind of situation)?

2) I've problem with the (known) parameters of these equations ( f_1 , f_2).

Let's see the matter:

the code is ( here there is a screen more clear because the subscripts make trouble):


Reduce[{(1/Sqrt[V (Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 3] - 
             V*MU)]) == (Subscript[f, 1]*Subscript[f, 2])^(-1/2),Sqrt[(Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 3] - V*MU)/
          V] Subscript[Z, 1]^-1 == (Subscript[f, 1]/Subscript[f, 2])^(
       1/2), Sqrt[(Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 3] - 
            V*MU)/V] Subscript[Z, 
        2]^-1 == (Subscript[f, 2]/Subscript[f, 1])^(1/2), 
      Sqrt[(Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 3] - V*MU)/
          V] Subscript[Z, 3]^-1 == (Subscript[f, 1]*Subscript[f, 2])^(
       1/2), ((Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 3] - V*MU)^(
         3/2)/(V^(3/2)*Subscript[Z, 1] *Subscript[Z, 2]*Subscript[Z, 
           3]) ) == (Subscript[f, 1]*Subscript[f, 2])^(1/2), MU == 0}, {V,
       Subscript[Z, 1] , Subscript[Z, 2], Subscript[Z, 3], MU}, Reals]
  • I have fixed the parameters to be positive but i obtain two conditional expressions associated to two values.

  • If you see it finds the value of Z_1 but it doesn't substitute its value into Z_3.

Thank you.

  • 5
    $\begingroup$ How much of this question is covered by: Solve vs Reduce? $\endgroup$ – Kuba Oct 2 '18 at 7:34
  • 5
    $\begingroup$ The subscripts will make debugging harder; reserve them only for display purposes. $\endgroup$ – J. M. is away Oct 2 '18 at 7:52
  • $\begingroup$ @Kuba I have written in the problem list if for this kind of situation is better Reduce or solve $\endgroup$ – Mordin Solus Oct 3 '18 at 13:25
  • $\begingroup$ @J.M.issomewhatokay. Before the stuff of the code there is the screeshot of the code which is the link called "code". $\endgroup$ – Mordin Solus Oct 3 '18 at 13:26
  • $\begingroup$ As I already said: the subscripts certainly look pretty, but they will make algebraic manipulation a pain. Regarding your use of MatrixForm[], see this. $\endgroup$ – J. M. is away Oct 3 '18 at 13:28

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