
Re: Simplifying Algebraic Expressions with Subtracted Expressions
Posted:
Nov 14, 2013 3:55 PM



On Nov 14, 2013, at 2:19 PM, Joe Niederberger <niederberger@comcast.net> wrote:
> Yes, I see that, but is that not, (more explicitly): > [1] (3/5)x  (1/4(13/5)x) = 200 ? > > Not exactly the same starting point, and the x really is there, done that way. Has to be. > > More explictly, the Ma version would be: > [2] x = 200 % (3/5  1/4(13/5)) > > But since the x is already isolated, I will grant the point that this is "arithmetic". But, although one can sweep that little first step that leads from [1 expression to Ma's under the rug of "done in the head", I think that would be a cheat.
I am saying that Ma's problem really isn't any different than the following problem that we do "in our head" ...
"Tom has 4 boxes and 200 donuts, how many donuts go in each box?"
This decodes to...
(4) Donuts Per Box = 200
In Ma's case, the problem decodes to...
(3/5  1/4(13/5)) Total Tarts Made = 200
The trick is reading the Ma problem directly into the coefficient (3/5  1/4(13/5)) in one pass, which seems foreign to us, but isn't really that hard, especially if you don't know algebra. So, in a sense, it is a harder reading problem but a simple arithmetic problem.
Yes, there is an implicit unknown in all problems, but, unless it requires algebra, then I don't call it "cheating" when you do these sorts of problems in "one step". How about 1.5 steps? To me, 2 steps implies that you have to look at the result of step 1 and make another decision, which is not the case here.
Bob Hansen

