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Re: Solvedoesn't

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  • Subject: [mg90642] Re: Solvedoesn't
  • From: Bill Rowe <readnews at>
  • Date: Thu, 17 Jul 2008 05:35:35 -0400 (EDT)

On 7/16/08 at 6:29 AM, fc3a501 at (Hauke Reddmann) wrote:

>This is extremely annoying. I have a very large (this seems to be
>critical) equation system. It's partly linear, partly horror, but
>already after I solve the linear part, a certain solution (which is
>obviously one since substituting it in makes all equations identical
>zero) can't be found anymore. The solution has some surds (this also
>seems critical, the ones without are found) but the equations are
>strictly polynomic.

>Faked Example (the real one only PM-ed, since it's a bit oversized):


>and after y was eliminated, it doesn't
>find x. (=golden mean here)

After fixing things to reflect correct syntax, Solve seems to
work just fine for the example you posted. That is,

In[5]:= eqns = {x - y == 1, x^2 + y^2 == 3, x^3 - y^3 == 4}=

In[6]:= Solve[eqns, {x, y}]

Out[6]= {{x -> (1/2)*(1 - Sqrt[5]), y -> (1/2)*(-1 - Sqrt[5])},
    {x -> (1/2)*(1 + Sqrt[5]), y -> (1/2)*(-1 + Sqrt[5])}}

Also, the first two equations are sufficient. That is,

In[7]:= Solve[Most@eqns, {x, y}]

Out[7]= {{x -> (1/2)*(1 - Sqrt[5]), y -> (1/2)*(-1 - Sqrt[5])},
    {x -> (1/2)*(1 + Sqrt[5]), y -> (1/2)*(-1 + Sqrt[5])}}

=46inally, I think Reduce is preferred when the number of
equations differs from the number of variables. And I like the
output I see better, i.e.,

In[8]:= Reduce[And @@ eqns, {x, y}]

Out[8]= (x == (1/2)*(1 - Sqrt[5]) ||
       x == (1/2)*(1 + Sqrt[5])) && y == x - 1

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