Unusual Simultaneous EquationsDate: 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/ |
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