Subtracting Different Types of Units

Date: 05/26/2001 at 16:41:16
From: Bruce Herman
Subject: Subtracting units of different types

I have a question about the following proposed solution for this 
question:

Tommy bought 20 pieces of candy for 20 cents. A piece of fudge costs 4 
cents, gumdrops are 4 for a penny, and chocolate drops are 2 for a 
penny. If he bought all these varieties, how many of each did he buy? 
	
Here is the solution one of my students came up with:

1)  4f + g/4 + c/2 = 20     multiply everything by 4

2) 16f + g   + 2c  = 80
3)   f + g   +  c  =  2     subtract line 3 from line 2
   --------------------
4) 15f       +  c  = 60

5) Since 15f is a multiple of 15, 15f can only be 15, 30, or 45, from 
   f = 1, 2, or 3 respectively.

6) If f = 1, then there are 45 chocolate drops; that's too many
7) If f = 2, then there are 30 chocolate drops; that's too many
8) If f = 3, then there are 15 chocolate drops; that's good

9) So we have 3 fudge and 15 chocolate drops, so we have 2 gumdrops, 
   which works out to 3 fudge = 12 cents; 15 chocolate drops = 
   7.5 cents; and 2 gumdrops = .5 cents; that does work.

The question the class has, which I can't explain, is that when we 
subtract line 3 from line 2, it looks as if we are subtracting the 
number of items from the cost of the items. How can we subtract when 
we are looking at different units?

Date: 05/26/2001 at 23:31:08
From: Doctor Peterson
Subject: Re: Subtracting units of different types

Hi, Bruce (and class!)

Interesting question. I have two ways to look at it.

First, in my mind when I write an equation it is abstract, and doesn't 
involve units. I hide the units in the definitions of the variables 
(for example, by saying "x is the length in meters," so that x itself 
is merely a number); in this case, "f is the number of pieces of 
fudge," and f is just a number, which becomes a number of pieces of 
fudge only when I apply my solution back to the actual problem. When 
you write the equation for cost, you are initially thinking of units,

     4 cents/piece * f pieces + 1/4 cent/piece * g pieces
     + 1/2 cent per piece * c pieces = 20 cents

but then you drop the units (dividing the whole equation by "cents") 
to get merely:

     4f + g/4 + c/2 = 20

Now when you do the algebra, you can ignore units; the algebra doesn't 
care.

Second, I have to admit there are times when I do like to think of an 
equation as having units, so that I can check, for instance, that 
every term represents an area, and I am not trying to add areas and 
lengths, which would indicate I made an error. If I want to do that 
here, I will have to say that equation (2) is in cents, while 
equation (3) is in pieces, and in order to subtract the latter I have 
to multiply it by the unit fraction "cents/piece" (just as I might 
have had to multiply it by 2 before subtracting) so that the units as 
well as the numbers match and I can really subtract like terms. I 
wouldn't bother with that sort of thinking here, but it might be worth 
considering.

Maybe there's a third way to think this out. Often equations like 
these can be solved concretely, as I have to do in solving this sort 
of problem with a student who hasn't had any algebra. That is, rather 
than work with variables, I would manipulate the actual objects, 
exchanging them, moving them to the other side of a scale, and so on. 
We would say in this case, first suppose Tommy bought four times as 
much of everything; then he would have spent 80 cents. (This gives 
equation 2.) Now suppose I pay him a penny for each item, regardless 
of its type (subtracting equation 3 from equation 2)... I'm not sure 
how I would finish this, but you might like to try it and see what 
concrete explanation you could give for that step. That would explain 
where the units go, and I suspect you will find it matches one of my 
answers above.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/27/2001 at 11:16:36
From: (Anonymous)
Subject: Re: Subtracting units of different types

How would that last work?  

        16f + g + 2c = 80
     -     f + g + c = 20

If each variable in the latter equation, as you suggest, is a penny, 
then how can you take away 20? Shouldn't you be taking away 3?

Date: 05/28/2001 at 22:56:53
From: Doctor Peterson
Subject: Re: Subtracting units of different types

Hi, Bruce.

Actually, this part doesn't make any sense to me either; it was late, 
I just had this idea that _some_ concrete solution might help explain 
what's going on, but I couldn't come up with a good solution along 
those lines, so I just left you with a suggestion of the kind of thing 
I had in mind, hoping you might be able to finish it, or that I would 
dream up such a solution. Let's think it through now.

The closest thing I can think of would be a concrete solution to 
something simpler, like:

    I bought 8 items, some for 5 cents and some for 2 cents, and
    they cost a total of 25 cents. What did I buy?

Algebraically, this becomes

      x + y = 8    (items)
    5x + 2y = 25   (cents)

and we subtract twice the first from the second to get

    3x = 9
     x = 3

     y = 8 - x = 5

To solve this without algebra, I would probably use an exchange 
method, like this:

    If all 8 items cost 2 cents, the total would be only 16 cents. I
    have to add 9 cents to that. If I replace one 2-cent item with a
    5-cent item, I have the same number of items but add 3 cents to
    the cost. In order to add 9 cents, I have to do this 3 times; so
    I will have 3 5-cent items and 5 2-cent items.

Algebraically, this is equivalent to substitution:

              y = 8 - x
  5x + 2(8 - x) = 25
        16 + 3x = 25
             3x = 9
              x = 3

The step that is equivalent to your addition of "counts and costs" is 
the substitution; but doing it this way, there's no unit conflict; I'm 
simply multiplying 2 cents/item by 8-x items. I think this suggests 
that what we're really doing when we subtract equations is multiplying 
the first equation not just by 2, but by 2 cents/item. That agrees 
with my second suggestion last time; but I don't think it really makes 
anything any clearer.

The fact is, combining two equations is an inherently abstract 
operation, and we're not really dealing with units at all, just 
numbers. The substitution method is more basic and intuitive, and fits 
better with the use of units.

I'll be interested to hear whether any of these ideas help your 
students at all!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/