Re: simultaneous equations for chemical speciation

*To*: mathgroup at smc.vnet.net*Subject*: [mg100222] Re: simultaneous equations for chemical speciation*From*: dh <dh at metrohm.com>*Date*: Thu, 28 May 2009 06:50:27 -0400 (EDT)*References*: <gvk9v0$ono$1@smc.vnet.net>

Hi Jem, First, the equation: a=b+c+d e=f+g+c+2d j=mg/f k=lm n+m+b+c=g+l do not add any information because they only introduce a new variable that does not appear anywhere else. This leaves 3 equations: h=c/(bg) i=d/(cg) hi=d/(bg^2) Multiplying the first 2 gives the third. Therefore, we actually have 2 equations: h=c/(bg) i=d/(cg) These do not have a general solution. Only under special circumstances we get a solution. For this problem we have Reduce. Therefore, you get possible solutions by: Reduce[{h == c/(b*g), i == d/(c*g)}, {g}] Daniel Jem & Tracy wrote: > Hi, I am a postgrad student and completely new to mathematica, I am just > starting to realise the possibilities. > > I am trying to do a chemical speciation calculation. > I have a set of simultaneous equations which represents the equilibria, > mass balance, and charge balance in the system. > > I want to eliminate some of the variables, and then be able to solve for > the remaining ones. > My system of equations looks like this: > a=b+c+d > e=f+g+c+2d > h=c/(bg) > i=d/(cg) > hi=d/(bg^2) > j=mg/f > k=lm > n+m+b+c=g+l > > Basically, I can solve for the variable I want, but I am having trouble > eliminating the ones I don't want (ie. I am getting an answer that is in > terms of the wrong variables) > Sofar my formula looks like: > > Solve[{a == b + c + d, e == f + g + c + 2 d, h == c/(b*g), > i == d/(c*g), h*i == d/(b*g^2), j == m*g/f, k == l*m, > n + m + b + c == g + l}, {g}] > > If I use a similar formula for the "Eliminate" function, I get error > mesages. > > I have been in the help files for ages, and don't really understand how > to do this. > I would really appreciate some pointers!! > Thanks > >