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### Variables Multiplied by Other Variables, Simplified Step by Step

```Date: 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.

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?

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.
```
Associated Topics:
High School Linear Algebra

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