Related quastions. Question 2

*To*: mathgroup at smc.vnet.net*Subject*: [mg57499] Related quastions. Question 2*From*: kazimir04 at yahoo.co.uk (Kazimir)*Date*: Sun, 29 May 2005 21:00:18 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

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