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. > >