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MathGroup Archive 2005

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Re: Solve with assumptions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58293] Re: Solve with assumptions
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 26 Jun 2005 01:33:55 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, England
  • References: <d86937$c7n$1@smc.vnet.net><d89261$s7f$1@smc.vnet.net> <d9isr2$cpk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Mukhtar Bekkali wrote:
> Great.
> 
> 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.
>

Hi Mukhtar,

*ToRules* is your friend in this case. For example,

In[1]:=
Reduce[x^2 - 1 == 0, x]

Out[1]=
x == -1 || x == 1

In[2]:=
ToRules[%]

Out[2]=
Sequence[{x -> -1}, {x -> 1}]

In[3]:=
{%}

Out[3]=
{{x -> -1}, {x -> 1}}

In[4]:=
x /. %

Out[4]=
{-1, 1}

Best regards,
/J.M.


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