Re: factor out a term to cancel in a fraction

*To*: mathgroup at smc.vnet.net*Subject*: [mg98887] Re: factor out a term to cancel in a fraction*From*: sean_incali at yahoo.com*Date*: Mon, 20 Apr 2009 19:14:31 -0400 (EDT)*References*: <200904200531.BAA00876@smc.vnet.net> <gshfsf$bvm$1@smc.vnet.net>

Initially, I thought the reason it didn't cancel was because maybe mathematica thought k1 could be = 0. So i figured, if I told it that k1>0, it may work. But now I see that Assuming probably has some other application... On Apr 20, 2:39 am, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote: > Factor[(e0*s[t]*k1*k2)/(s[t]*k1 + k1*Km)] > (e0*k2*s[t])/(Km + s[t]) > > or > > Cancel[(e0*s[t]*k1*k2)/(s[t]*k1 + k1*Km)] > (e0*k2*s[t])/(Km + s[t]) > > Difficult ? Give an example of "easy". > > And by the way, where did you get the idea that Assuming does anything = > at all with Collect? > > Andrzej Kozlowski > > On 20 Apr 2009, at 14:31, sean_inc... at yahoo.com wrote: > > > Hello group. > > > I'm trying to do a little algebra which i can do in my head using > > Mathematica. Why? Well, it's neater. > > > Consider the following. > > > p'[t] == (e0 s[t] k1 k2 )/ (s[t] k1 + k1 Km) > > > Factor out k1 and cancel it to simplify the expression. Why is this so > > difficult to accomplish in mathematica? Following does nothing. > > > p'[t] == (e0 s[t] k1 k2 )/ (s[t] k1 + k1 Km) // > > Assuming[k1 > 0, {Collect[#, k1]}] & > > > Thanks for any info. > > > Sean

**References**:**factor out a term to cancel in a fraction***From:*sean_incali@yahoo.com