Subtracting Different Types of UnitsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/