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Re: Re[2]: cannot solve *trivial* equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg34421] Re: Re[2]: [mg34380] cannot solve *trivial* equation
  • From: Andrzej Kozlowski <andrzej at platon.c.u-tokyo.ac.jp>
  • Date: Sun, 19 May 2002 04:14:36 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On Friday, May 17, 2002, at 09:49  PM, Marco Manfredini wrote:
>
> I [=Mathematica newbie] did not understand that Mathematica can solve 
> for A {A->3*S} but
> not for S ({} while I expected S->A/3). I had the idea that solving
> for X means to find an equation X==Y (with Y X-free) which can be
> added to the set of equations without restrictions.
>
Note that your two examples are interpreted differently by Mathematica. 
In the first case:

Solve[{A == S + Q, Q == 2*S}, A]

Q == 2*S is a relation between the parameters (not variables) which 
mathematica takes as assumption. But in the second case:

Solve[{A == S + Q, Q == 2*S}, S]

there is now assumption made about the parameters. So of course if you 
add to the equations one more condition:

In[8]:=
Solve[{A == S + Q, Q == 2*S, 2*A == 3*Q}, S]

Out[8]=
{{S -> Q/2}}

You get the answer you were expecting. In fact, if you prefer to get the 
answer it in terms of A you should use:

In[11]:=
Solve[{A == S + Q, Q == 2*S, 2*A == 3*Q}, S, {Q}]

Out[11]=
{{S -> A/3}}


Andrzej


Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/



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