I am trying to solve a 3 unknowns 3 equations system. The code does not give any error but gives any output neither.

Solve[ {zj zu (zo + \[Phi]1 no) - a3 nj zo zu \[Phi]1 + zj (a1 no zu \[Phi]1 + zo (zu + a2 nu \[Phi]1)) == 0, zo zu (zj + \[Phi]1 nj) - a3 nj zo zu \[Phi]1 + zj (a1 no zu \[Phi]1 + zo (zu + a2 nu \[Phi]1)) == 0, zj zo (zu + \[Phi]1 nu) - a3 nj zo zu \[Phi]1 + zj (a1 no zu \[Phi]1 + zo (zu + a2 nu \[Phi]1)) == 0}, {a1, a2, a3}]

I appreciate any hint, solution or suggestion for this issue.

  • 4
    $\begingroup$ Try Reduce instead of Solve. It looks like there exists no general solution to this system of equations. $\endgroup$ – Sjoerd Smit Oct 27 at 14:44
  • 2
    $\begingroup$ The output of { } is an output. It indicates that there is no general solution (see documentation for Solve). $\endgroup$ – Bob Hanlon Oct 27 at 14:50
  • 1
    $\begingroup$ Try giving conditions for you unspecified symbols like e.g. nj>0,... $\endgroup$ – Daniel Huber Oct 27 at 14:52
  • $\begingroup$ With red = Reduce[eqs, {a1, a2, a3}] you get a lot of cases where equations are satisfied, if some of the parameters are zero and others not. In that case all variables ai and the other parameters can take every value. If you don't allow any parameter to be zero red2 = Solve@ Reduce[Join[eqs, Thread[{nj, no, nu, zj, zo, zu, \[Phi]1} != 0]], {a1, a2, a3}] you get the only solution for a variable depending on the other variables and some parameters: {{nu -> (no zu)/zo, zj -> (nj zo)/no, a1 -> (-2 zo - no \[Phi]1 - a2 no \[Phi]1 + a3 no \[Phi]1)/( no \[Phi]1)}} $\endgroup$ – Akku14 Oct 28 at 19:54

If you solve your first equation by hand for a3, enter this into the other two equations and expand all sides your system collapses to

no zj == nj zo 

which doesn't contain any ai's any more...

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