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Re: Solve / Reduce isolating results.

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Hi,

first, in view of simplification, reduction and solving it's very nice
for Mathematica to have exact numbers if you can provide them. In this
case, you can. Either you know the exact values (e.g. 2.75 == 11/4) or
you let Mathmatica find it out. Then use Reduce and give the domain
Reals and your other wishes and everything is fine:

eq = {(1/6)*((SQ1*196)*2)^3 - (1/
        6)*((((2.75 + XX)*((3.75 + XX)*21^2))*100)*2)^3 == 
    1.2902698935932682*^18, (1/6)*((SQ1*100)*2)^3 - (1/
        6)*((((2.75 + XX)*((3.75 + XX)*21^2))*36)*2)^3 == 
    1.883856772422697*^17} /. 
  x_ /; Head[x] =!= Symbol && ExactNumberQ[x] === False :> 
   Rationalize[x, 0]
Reduce[Flatten[{eq, SQ1 > 0, XX > 0}], {SQ1, XX}, Reals]

and you get

Out[13]= SQ1 == 5292 && XX == 1/4

Cheers
Patrick

On Fri, 2011-04-29 at 07:35 -0400, Lea Rebanks wrote:
> Dear All,
> 
> Given the enclosed code - see below.
> Please could someone help me setup the code to Solve / Reduce for these 2
> missing values only required 
> 
>   {SQ1 -> 5292, 
>     XX -> 0.25}
> 
> (I don't want multiple results.) I want positive numbers & NOT Complex..
> 
> 
> Many thanks for your help & attention, really appreciated. Please see below.
> 
> Best regards,
> Lea...
> ...................................................................
> 
> 
> Clear[SQ1]
> Clear[XX]
> Solve[{(1/6)*((SQ1*196)*2)^3 - 
>          (1/6)*((((2.75 + XX)*((3.75 + XX)*21^2))*100)*2)^3 == 
>        1.2902698935932682*^18, (1/6)*((SQ1*100)*2)^3 - 
>          (1/6)*((((2.75 + XX)*((3.75 + XX)*21^2))*36)*2)^3 == 
>        1.883856772422697*^17}, {SQ1, XX}]
> 

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Re: Solve / Reduce isolating results. Patrick Scheibe <pscheibe@trm.uni-leipzig.de> - 2011-04-30 09:52 +0000

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