Variables Multiplied by Other Variables, Simplified Step by StepDate: 08/22/2011 at 23:28:28 From: David Subject: Solve system of equations where variables are multiplied How can I solve a system of equations where some of the variables are multiplied together? or determine if a solution is even possible? My original problem has 22 equations and 23 variables, of which one variable is given. For brevity's sake, here's an example of a simpler problem. Given D, solve for F: D = B*E E = C*F G = B + D A = C + E C = F + D - 1 B = G*C - 1 F = A + G + 1 Some variables appear in more than two equations. Overall, I count seven equations and seven variables. I also notice the number of times each variable appears: A appears 2 times B appears 3 times C appears 4 times D appears 3 times E appears 3 times F appears 3 times G appears 3 times But due to multiplied variables, I cannot solve these simultaneous equations in the standard way (matrix reduction); and substitution leads to null equations that zero out on both sides. Can you tell by the number of equations and variables whether the problem is even solvable? Does it matter if variables appear an odd number or even number of times in the system? What is the approach for solving this type of system? Date: 08/23/2011 at 01:35:43 From: Doctor Vogler Subject: Re: Solve system of equations where variables are multiplied Hi David, Thanks for writing to Dr. Math. There are general strategies for solving a system of polynomial equations. Sometimes the equations have dependencies among them, but usually you can use one equation to eliminate one variable, a process you can repeat until you have just one variable left (or just one equation left). In the case of 22 equations in 23 variables, you can do this until you have one equation in two variables, and you would probably want to do this so that the two variables are the "given" one and the one you want to solve for. In the case of 7 equations in 7 variables, you can do this until you have one equation in one variable, which means that there will only be finitely many solutions at all, so you don't generally need a given. (You could use the "given" variable to distinguish among the different solutions, as long as you realize that the "given" variable has only finitely many values that will solve the equations.) Usually it is best to eliminate a variable when you can easily solve for it and it equals something simple, because each time you substitute its value into the other equations, they all become more complicated (such as higher degree polynomials). The order in which you eliminate the variables can make a big difference in how easy it is to solve the equations, but this is more art than science. You are welcome to try things one way, and then if it gets ugly, you can try another way and see if it goes more smoothly. If you have lots of equations and variables, then sometimes there is no way to avoid very messy equations. Computer algebra programs (like Mathematica, Maple, GNU Pari, Magma, and others) can help with this. For example, let's start with your equations: D = B*E E = C*F G = B + D A = C + E C = F + D - 1 B = G*C - 1 F = A + G + 1 First, eliminate A. We have A = C + E, which we substitute into the other equations: D = B*E E = C*F G = B + D C = F + D - 1 B = G*C - 1 F = (C + E) + G + 1 Next, eliminate B, which is B = G*C - 1: D = (G*C - 1)*E E = C*F G = (G*C - 1) + D C = F + D - 1 F = (C + E) + G + 1 Next, eliminate C, which is C = F + D - 1: D = (G*(F + D - 1) - 1)*E E = (F + D - 1)*F G = (G*(F + D - 1) - 1) + D F = ((F + D - 1) + E) + G + 1 Things are getting messier, but we have fewer equations and variables. Now, eliminate E, which is E = (F + D - 1)*F: D = (G*(F + D - 1) - 1)*((F + D - 1)*F) G = (G*(F + D - 1) - 1) + D F = ((F + D - 1) + ((F + D - 1)*F)) + G + 1 Okay, now we'd better clean up a bit and see where we are: D = (G*(F + D - 1) - 1)*(F + D - 1)*F G = G*(F + D - 2) + D F = F + D - 1 + (F + D - 1)*F + G + 1 Now we can solve the last equation for G, since G only appears once. We get G = -D - (F + D - 1)*F. We are left with two equations: D = ((-D - (F + D - 1)*F)*(F + D - 1) - 1)*(F + D - 1)*F G = (-D - (F + D - 1)*F)*(F + D - 2) + D Here, we can't easily solve for any more variables. But we can put these two equations together to reduce their degree in D. Currently, they have degrees three and two in D. We multiply the second equation by D*F and subtract the two equations, giving us another equation with degree two in D. Then we multiply that by F + 1 and subtract that result from the second equation above times F^3 + F^2 + F. The result has degree one in D, so we can solve for D: D = (-3*F^3 + 4*F^2 + 2*F)/(5*F^2 + 5*F + 1) When we substitute that into the second equation, we find that F must satisfy the polynomial 4*F^7 + 4*F^5 + 9*F^4 + 6*F^3 + 10*F^2 + 6*F + 1 = 0. This factors as (4*F^4 - 4*F^3 + 4*F^2 + 5*F + 1)(F^2 + 1)(F + 1) = 0. But actually, if F = -1, then the second equation says D = 5, which is not a solution to the first equation. That means that F has to satisfy the equation (4*F^4 - 4*F^3 + 4*F^2 + 5*F + 1)(F^2 + 1) = 0. That equation has four complex solutions (including i and -i) but only two real solutions, which are approximately F = -0.3003688680356141425729479754 F = -0.4013776088512037125466878061 The variable F can be any of those six solutions. Now, since you have F, you can back-solve for the other variables: D = (-3*F^3 + 4*F^2 + 2*F)/(5*F^2 + 5*F + 1) G = -D - (F + D - 1)*F E = (F + D - 1)*F C = F + D - 1 B = G*C - 1 A = C + E You could also try eliminating the variables in the order A, G, C, B, E, D. How does that change things? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 08/23/2011 at 20:17:29 From: David Subject: Thank you (Solve system of equations where variables are multiplied) Thank you so much. The example really helps me understand what I am dealing with and how messy it can get with the substitutions. Showing how to reduce the degree also is very helpful. Thank you again. |
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