Re: Solving for multiple variables in ONE equation.

Randy,

Ah ha.. Makes perfect sense now. That's what I get for skimming and
typing right before I head for home.

While some of the B's could hypothetically be zero they usually aren't,
so it looks like an infinite number of possible solutions which is more
or less what I expected when I asked the question.

Hmm... Ok, looks like we'll be going back to the drawing board on some
of this, but at least now I can speak from a position of semi-authority
about it. :)

Thanks again!

Robert

Randy Poe wrote:
rgam...@xxxxxxxxx wrote:
Hagman (and others, but Hagman specifically has a comment I want to
find out more about),

First of all, thanks for the prompt replies!

I'm unsure exactly what you mean by the last part of your reply,
specifically:

"
In the equation you gave, it is always possible to replace Z2 by Z2+B3
and Z3 by Z3-B2 to obtain a new solution.
"

So we would end up with:

X = (Y*Z1) / ( (B1*Z1 + (B2 * (Z2+B3)) + (B3 * (Z3-B2)) ) ?

Remember, my math is somewhat rusty, so I'm not sure I understand the
value of this? Or is the intent to create three equations that can be
then solved simultaneously?

Hagman is just pointing out that given a solution, there is
at least one other solution known immediately. For instance
if Z1=1, Z2=2, Z3=3 is a solution, then so is Z1=1, Z2=2+B3,
Z3=3-B2.

The reason that is so is that

B2*(Z2+B3) + B3*(Z3-B2) = B2*Z2 + B2*B3 + B3*Z3 - B3*B2
= B2*Z2 + B3*Z3

So using these two new values doesn't change the
denominator, and the whole fraction still equals X.

More than that, if you add r*B3 to Z2 and subtract
r*B2 from Z3, you'll get another valid solution for any
value of r for the same reason (it doesn't change the
denominator).

As I pointed out, in general if your B's are nonzero, there
are infinitely many possible solutions, and you can
choose any (n-1) of your n variables to be anything
you want except all 0's.

- Randy

.

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