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MathGroup Archive 2005

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Re: Solve with assumptions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58284] Re: Solve with assumptions
  • From: Bill Rowe <readnewsciv at earthlink.net>
  • Date: Sun, 26 Jun 2005 01:33:47 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 6/25/05 at 1:56 AM, mbekkali at gmail.com (Mukhtar Bekkali) wrote:

>How do you pull the solution out from Reduce? That is, with Solve I
>can have x=x/.Solve[f[x]==0,x] and get x and use it later on.  The
>output of Reduce is identity, x==number.

Keep in mind x==number is an expression which can be manipulated with other Mathematica functions.

Specifically, note

FullForm[Reduce[x + 1 == 0,  x]]

results in 

Equal[x, -1]

So, the -1 can be extracted using Apply as follows:

Last[List@@Reduce[x + 1, x]]

or even more directly with Part as

Reduce[x + 1 == 0, x][[2]]

or

Part[Reduce[x +1 == 0, x], 2]
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