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Re: Re: Solve with assumptions

• To: mathgroup at smc.vnet.net
• Subject: [mg58285] Re: [mg58277] Re: Solve with assumptions
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sun, 26 Jun 2005 01:33:48 -0400 (EDT)
• Reply-to: hanlonr at cox.net
• Sender: owner-wri-mathgroup at wolfram.com

soln=Reduce[b*x+c==0,x]

(c == 0 && b == 0) || (b != 0 && x == -(c/b))

x/.ToRules[Last[soln]]

-(c/b)

soln[[2,2,2]]

-(c/b)


Bob Hanlon

> 
> From: "Mukhtar Bekkali" <mbekkali at gmail.com>
To: mathgroup at smc.vnet.net
> Date: 2005/06/25 Sat AM 01:56:38 EDT
> Subject: [mg58285] [mg58277] Re: Solve with assumptions
> 
> 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.
> 
>