FAQS/follow-up on solving a set of symbolic equations

There are two "standard" ways of doing what you want,
namely accessing the result of solve.

One is to do  ev(a,d5);
or just a,d5;
or the subst given by Martin.

the other way is by setting some flag
globalsolve:true;

which SETS a to its value.


Regarding the Ci Di discussion:  yes this has
been a source of user misunderstanding, bugs etc.

C= command, D= display  E= extra  (I think).
They can be reset as mentioned.
It is also possible to refer to the last expression
as %, and maybe previous ones as %[2], %%,  ...
This was copied by Mathematica, and I think Maple,
and Mupad copied it from Maple, I think.
Anyway, In[1], Out[1]  is very clumsy to type,
so there is a tradeoff.

RJF

RJF


Martin RUBEY wrote:
> 1. both the inchar, outchar and the answer below should make it into the 
> faq, I think.
> 
> 2. use subst:
> 
> GCL (GNU Common Lisp)  Version(2.5.0) Sun Nov 17 15:58:09 CET 2002
> Licensed under GNU Library General Public License
> Contains Enhancements by W. Schelter
> Maxima 5.9.0rc3 http://maxima.sourceforge.net
> Distributed under the GNU Public License. See the file COPYING.
> Dedicated to the memory of William Schelter.
> This is a development version of Maxima. The function bug_report()
> provides bug reporting information.
> (C1) a + b*c=1;
> 
> (D1) 				  b C + a = 1
> (C2) b - a*c=0;
> 
> (D2) 				  b - a C = 0
> (C3) a+b=5;
> 
> (D3) 				   b + a = 5
> (C4) display2D:false;
> 
> (D4) FALSE
> (C5) solve([d1,d2,d3],[a,b,c]);
> 
> (D5) [[a = (25*SQRT(79)*%I+25)/(6*SQRT(79)*%I-34),
>        b = (5*SQRT(79)*%I+5)/(SQRT(79)*%I+11),C = (SQRT(79)*%I+1)/10],
>       [a = (25*SQRT(79)*%I-25)/(6*SQRT(79)*%I+34),
>        b = (5*SQRT(79)*%I-5)/(SQRT(79)*%I-11),C = -(SQRT(79)*%I-1)/10]]
> (C6) subst(d5[1],a);
> 
> (D6) (25*SQRT(79)*%I+25)/(6*SQRT(79)*%I-34)
> (C7) 
> 
> 
> 
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