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Re: Solving an equation
*To*: mathgroup at smc.vnet.net
*Subject*: [mg67500] Re: Solving an equation
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Wed, 28 Jun 2006 03:52:38 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <e7qmrt$6oa$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
masha wrote:
> Hi,
>
> I need to solve an equation (below), however, when I use Solve, I get
> an answer which I know cannot be correct (probably because Mathematica
> uses Inverse). Is their a way to get around this problem and get a true
> answer?
>
> \!\(Assuming[\ A > 0\ && \ \
> B\ > A\ && \ b > 1\ && \(P\^b\ Q\)\/a\ < \ \ B\ && \ \(P\^b\
> Q\)\/a \
>> A\ && \ P\^b\ Q = a\ B && \ a\ > \ 0\ && \ Co\ > \ 0\ && \ Cu\ >
> 0\ && tr\ > 0\ && \ tm > 0\ && \ Q > 0 && \ P >
> 0\ && T > 0, Solve[\(P\^\(\(-1\) - b\)\ \((\(-2\)\ a\ B\ P\^\(1
> + b\)\
> Q + P\^\(2\ b\)\ \((P + b\ \((Co + Cu + P)\))\)\ Q\^2 - a\^2\ \((
> A\^2\ b\ \((Co + T)\) + 2\ A\ \((\((\(-1\) + b)\)\ P - b\ Q\
> T)\) + B\ \((\(-\((\(-2\) + B)\)\)\ P + b\ \((\(-2\)\ P +
> B\ \((Cu +
> P - T)\) + 2\ Q\ T)\))\))\))\)\)\/\(2\ a\ \((A - B)\)\) == 0,
> P]]\)
>
> Thank you,
> Masha
>
Hi Masha,
Your system seems to be inconsistent since (P^b*Q)/a < B && P^b*Q == a*B
==> B < B.
You should eliminate the inequalities tr > 0 && tm > 0 since the
variables tr and tm are not used in the equation.
It is best to use a replacement rule {for instance /.P^b*Q -> a*B) or a
With clause [1] (for instance With[{B = P^b*Q/a}, ... ]) in place of the
equality P^b*Q == a*B.
Finally, try using the built-in function Reduce [2] rather than Solve.
HTH,
Jean-Marc
[1] http://documents.wolfram.com/mathematica/functions/With
[2] http://documents.wolfram.com/mathematica/functions/Reduce
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