quasi
Posts:
11,458
Registered:
7/15/05


Re: Solving a system of multivariable equations with first order variables
Posted:
May 2, 2014 6:49 PM


Ron wrote: > >Hello, are there standard methods for solving systems of >multivariable 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

