Re: Related quastions. Question 2

*To*: mathgroup at smc.vnet.net*Subject*: [mg57524] Re: [mg57499] Related quastions. Question 2*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Tue, 31 May 2005 04:59:27 -0400 (EDT)*Reply-to*: hanlonr at cox.net*Sender*: owner-wri-mathgroup at wolfram.com

Using a name does not change the form of the expression expr=a x+b==c expr1=#-b&/@expr expr2=#/a&/@expr1 b + a*x == c a*x == c - b x == (c - b)/a Bob Hanlon > > From: kazimir04 at yahoo.co.uk (Kazimir) To: mathgroup at smc.vnet.net > Date: 2005/05/29 Sun PM 09:00:18 EDT > Subject: [mg57524] [mg57499] Related quastions. Question 2 > > Dear Bob, > > you gave the example > > Print["Equation to solve for x"] > a x + b == c > Print["Subtract b from each side"] > # - b & /@ %% > Print["Divide each side by a"] > #/a & /@ %% > > It works but i do not like very much expressions with %%, as I should > not insert more lines before it or I have to modify this symbol. I > would like to start with > > expr= (a x + b == c) > > and then make a transform like expr1 = expr - b. It clearly does not > work. Could you advise me a mathematical operartion over expression > which would return a x == c - b . > > The best thing I found was > expr = (a x + b == c) > Distribute[#1 - b &@expr, Equal] > however, it does not work in all cases that I need. > > Vlad > >