On Mon, Nov 14, 2011 at 17:56, Richard Hennessy
<rich.hennessy at verizon.net>wrote:
> This will not work. The number of equations has to match the length of
> the list of variables to solve.
>
That is not quite true. There can be more variables than equations, in
which case arbitrary numbers may be part of the solution:
%i10) linsolve([a=b,b=c],[a,b,c]);
(%o10) [a = %r4,b = %r4,c = %r4]
There can be fewer variables than equations, in which case some of the
equations must be redundant for there to be a solution:
(%i13) linsolve([a=b,b=c,a=c],[a,b]);
solve: dependent equations eliminated: (3)
(%o13) [a = c,b = c]
But the solutions must be true for *all* values of variables which are not
being solved for -- you can call those variables "parameters". For
example, the following has zero solutions
(%i14) linsolve([a=b,b=c,x=3],[a,b]);
(%o14) []
because there is no way x=3 can be made true for all x.
-s