Re: Solve output depends on previous attempt with bad syntax
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- Subject: [mg132513] Re: Solve output depends on previous attempt with bad syntax
- From: "djmpark" <djmpark at comcast.net>
- Date: Fri, 4 Apr 2014 03:57:38 -0400 (EDT)
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Clear[ y1, y2] David Park djmpark at comcast.net http://home.comcast.net/~djmpark/index.html From: Alain Cochard [mailto:Alain.Cochard at unistra.fr] First, the correct solution of a linear system of 2 equations with 2 unknowns, for reference. Mathematica 9.0 for Linux x86 (64-bit) Copyright 1988-2013 Wolfram Research, Inc. In[1]:= Solve[{y1==a x1 +b, y2==a x2 +b},{a,b}] -y1 + y2 x2 y1 - x1 y2 Out[1]= {{a -> -(--------), b -> -(-------------)}} x1 - x2 x1 - x2 Now, starting a new Mathematica session, assume I make a mistake, using '=' instead of '==', I get an error message -- so far, so good: Mathematica 9.0 for Linux x86 (64-bit) Copyright 1988-2013 Wolfram Research, Inc. In[1]:= Solve[{y1=a x1 +b, y2=a x2 +b},{a,b}] Solve::naqs: b + a x1 && b + a x2 is not a quantified system of equations and inequalities. Out[1]= Solve[{b + a x1, b + a x2}, {a, b}] Realizing my mistake, I retry with the proper syntax: In[2]:= Solve[{y1==a x1 +b, y2==a x2 +b},{a,b}] Out[2]= {{}} which is obviously not the correct result. Is there a rationale here, i.e., could I have anticipated this output? Thank you, Alain -- EOST (=C3=83=E2=80=B0cole et Observatoire des Sciences de la Terre) IPG (Institut de Physique du Globe) | alain.cochard at unistra.fr 5 rue Ren=C3=83=C2=A9 Descartes [bureau 106] | Phone: +33 (0)3 68 85 50 44 F-67084 Strasbourg Cedex, France | Fax: +33 (0)3 68 85 01 25