On Tuesday, February 8, 2011 1:42:38 PM UTC-7, Nils Bruin wrote: > > On Feb 8, 10:30 am, tvn <[email protected]> wrote: > > this is what I am trying to do -- is there a way to have it to return uk > = > > w-y ? > [...] > > sage: solve([uk + x + 2*y == A*B, w + x + y == A*B],[uk,x]) > > [[uk == w - y, x == A*B - w - y]] > > You answer your own question. >
Not entirely , I was able to obtained uk == w - y by asking it to solve for [uk,x] ... if I ask it to solve for [uk, v] where v is some other variables in the provided equations and uk == might be something else. In other words, I 'guessed' x and it was right. -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
