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MathGroup Archive 2005

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Re: Don't understand behaviour of Solve[]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58587] Re: Don't understand behaviour of Solve[]
  • From: dh <dh at metrohm.ch>
  • Date: Sat, 9 Jul 2005 04:07:48 -0400 (EDT)
  • References: <daitlm$sus$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Oliver,
Mathematica can not know what you consider to be a parameter and what a
variable. If you do not specify a variable Mathematica takes it for a
parameter. But then you have an overdetermined system without a generic 
solution. If you want the solution only for a fraction of variables you 
have to tell Mathematica to eliminate the rest by specifying a third argument.
Example:
Solve[{x + y == 3, x + 2 y == 5}, {x, y}]  gives {{x -> 1, y -> 2}}
Solve[{x + y == 3, x + 2 y == 5}, {x}] gives {} because Mathematica takes y now 
as a parameter, not  a variable. You have 2 equations for 1 variable 
that do not have a generic solution. A solution only exists for a 
special value of y.
If you only want a solution for x, you must tell Mathematica by the third
argument:
Solve[{x + y == 3, x + 2 y == 5}, {x}, y]  gives {{x -> 1}}

sincerely, Daniel


Oliver Friedrich wrote:
> Hallo,
> 
> I have a problem understanding the general behaviour of the Solve[] 
> function.
> 
> I have a set of equations with the variables a,b,c,d and k.
> 
> Evaluating Solve[set,k] or Solve[set,{a,b}] returns with {}, whereas Solve
> [set,{k,b,c,d}] returns solutions for k,b,c and d.
> 
> My question: Why does Solve returns no solution when searching only for k 
> for example and why is there a solution for several variables? I thought 
> that giving a list of variables just means that I want to search for all of 
> them. But the procedure doesn't seem to be independant from one variable to 
> another.
> 
> What kind of information contains my list of variables except my wish to 
> solve for these?
> 
> Many thanks
> 
> Oliver friedrich
> 


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