Re: Solve / Reduce isolating results.
- To: mathgroup at smc.vnet.net
- Subject: [mg118513] Re: Solve / Reduce isolating results.
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 30 Apr 2011 05:53:23 -0400 (EDT)
eqns = {(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} // Rationalize[#, 0] &;
Solve[eqns, {SQ1, XX}, Reals]
{{SQ1 -> 5292, XX -> -(27/4)}, {SQ1 -> 5292, XX -> 1/4}}
Select[Solve[eqns, {SQ1, XX}], Element[#[[All, -1]], Reals] &]
{{SQ1 -> 5292, XX -> -(27/4)}, {SQ1 -> 5292, XX -> 1/4}}
Bob Hanlon
---- Lea Rebanks <lrebanks at netvigator.com> 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}]