Hi,
Am Montag, den 15.03.2010, 17:51 -0700 schrieb Mike:
> var('x,fa,fb,fc')
> solve ([(43*2.5)==((fb*(5-x))+(fa*5)), (43*(2.5-x))==(fc*(5-x)),
> (43*2.5)==((fb*x)+(fc*5)), 43==(fa+fb+fc)], x,fa,fb,fc)
from sage: help(solve):
> If there is a parameter in the answer, that will show up as
> a new variable. In the following example, ``r1`` is a real free
> variable (because of the ``r``)::
>
> sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
> [[x == -r1 + 3, y == r1]]
Your equations don't determine all your variables, there is a "free" on, named
"r...".
Eckhard
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