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### Subtracting Different Types of Units

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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

- 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.

students at all!

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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