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Unusual Simultaneous Equations

Date: 02/06/2003 at 15:01:43
From: Khursheed
Subject: Solving an Equation

I don't know how to solve this. I have

A = a*(b+c)/(a+b+c)
B = b*(a+c)/(a+b+c)
C = c*(a+b)/(a+b+c)

A, B, and C are known. I don't know how to solve for a, b, and c.

I just need a little direction. I tried to put it in matrix form to 
reduce it somehow but I cannot reduce it to one equation where a, b, 
and c can be expressed in terms of A, B, and C alone.

I have set a+b+c = s  so wherever there was a+b I subsituted (s-c) and
so on. Yet whenever I multiply, it becomes a product of at least two
variables in every equation. I just don't know how to separate them. 



Date: 02/09/2003 at 07:34:41
From: Doctor Jacques
Subject: Re: Solving an Equation

Hi Khursheed,

This is a tough one indeed.

Notice that, in your equations, if we multiply a, b, and c by the 
same number, A, B, and C will also be multiplied by that number, and 
the equations will have the same form.

This suggests (but this nothing more than a guess) that we concentrate 
on the proportion (a : b : c).

Let us write s = a + b + c, as you did.

Subtract the second equation from the first :

  s(A - B) = c(a - b)

and compare with the third:

  sC = c(a + b)

this should give us the ratio (a-b)/(a+b), and therefore, also the 
ratio (a/b).

With a little algebra, you should find:

  b(A-B+C) = a(-A+B+C)

Using the same trick with other equation pairs (or simply using 
symmetry), we find

  a(-A+B+C) = b(A-B+C) = c(A+B-C)

Let us rewrite this as:

  (a/P) = (b/Q) = (c/R) = k,

where P = 1/(-A+B+C), etc., and k is the (unknown) common value of 
these expressions.

P, Q, and R are known, we still have to compute k.

Pick one of the equations (for example the first), and substitute:

  a = Pk, b = Qk, c = Rk

This will give you an equation in k; knowing k, you can find a, b, c 
by the equations just above.

You will note that the equation in k, when written using P, Q, and R, 
does not look symmetrical. However, if you write P, Q, R as functions 
of A, B, and C (we would not do that in practice), some 
simplifications will happen (for example, Q+R has A as a factor), and 
the final result is symmetrical in A, B, C as it should be.

To allow you to check your intermediate calculations, I worked out an 
example:

Let a = 1, b = 2, and c = 3 (I started from the solution, but let us 
forget it...)

You should find:

  A = 5/6, B = 4/3, C = 3/2

This is the statement of the problem.

Using the method above, you should find:

  P = 1/2
  Q = 1
  R = 3/2
  k = 2

and this gives back the solution we started from.

Note that there may be a lot of special cases, which you will 
recognise when you try to divide by 0.

Example 1
----------

Let us assume that A = B

The first equation we wrote: s(A-B) = c(a-b) shows that either c = 0, 
or a = b.

If c = 0, we must have C = 0; otherwise a = b and we must have A = B 
(=0). If neither is the case, the system has no solution.

If A=B=0 and C<>0, you should use the third original equation in the 
last step to compute k. (It is even easier to start back at the 
original equations).

Example 2
---------

Let us assume that A = B+C. In this case, P is not defined.

If we look at the original equation, we find that, in such a case, we 
find that bc=0, and therefore, either C = 0 (and A = B) or B = 0 (and 
A = C). If neither is true, there are no solutions; otherwise, it is 
much easier to solve the remaining system.

I hope this helps; please write back if you require further help.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra

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