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.