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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