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Topic: Solving a system of multi-variable equations with first order variables
Replies: 3   Last Post: May 2, 2014 7:13 PM

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 quasi Posts: 12,067 Registered: 7/15/05
Re: Solving a system of multi-variable equations with first order variables
Posted: May 2, 2014 6:49 PM

Ron wrote:
>
>Hello, are there standard methods for solving systems of
>multi-variable equations where all variables are of order one,
>such as the following?:
>
>c11.xyz + c12.xy + c13.yz + c14.xz + c15.u + c16.v + c17.w = c10
>c21.xyz + c22.xy + c23.yz + c24.xz + c25.u + c26.v + c27.w = c30
>c31.xyz + c32.xy + c33.yz + c34.xz + c35.u + c36.v + c37.w = c30
>
>There are always an equal number of equations and variables,

In the example above, there are 3 equations and 6 variables.

Were some equations omitted?

Note that while the equations are degree 1 in each variable,
the total degree is 3 (since the term xyz is cubic).

Assuming there are 3 omitted equations (for a total of 6),
I would treat each of the terms u,v,w,xy,yz,zx,xyz as
separate variables and solve the resulting underdetermined
system of linear equations (6 equations, 7 unknowns). In
general, the solution will have a free parameter, t say,
and can be expressed in the form

u = (a1) + (b1)*t
v = (a2) + (b2)*t
w = (a3) + (b3)*t
xy = (a4) + (b4)*t
yz = (a5) + (b5)*t
zx = (a6) + (b6)*t
xyz = (a7) + (b7)*t

where a1,...,a7 and b_1,...,b_7 are now known constants, and
the variable t is still free.

Then using relation

(xy)*(yz)*(zx) = (xyz)^2

you get an equation in t of degree at most 3, allowing you
to solve for t.

Of course, there may be as many as 3 solutions.

For each valid t, you now have values for

u,v,w,xy,yz,zx,xyz

Finally, solve for x,y,z using the relations

x = (xyz)/(yz)
y = (xyz)/(zx)
z = (xyz)/(xy)

and the problem is solved.

quasi

Date Subject Author
5/2/14 Ron
5/2/14 quasi
5/2/14 Ron
5/2/14 Ron