       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

>

>

```

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