Solving Literal EquationsDate: 09/08/2001 at 14:15:31 From: Jenny Subject: Literal equations I just don't get how to do literal equations. I can do equations fine with just numbers, but doing them literally with letters is confusing me. Do you think that you can explain literal equations to me, please? Thanks! Date: 09/08/2001 at 22:43:18 From: Doctor Peterson Subject: Re: Literal equations Hi, Jenny. I'd like to see just how you are confused, so I can help better. The explanation may not be satisfying to you: You do exactly the same thing to solve a "literal equation" as for one with only one variable. In fact, you do a little less, since you can't actually carry out multiplications and additions, but just simplify instead. (Once you know algebra well enough, this is so commonplace that the term "literal equation" seems unnecessary; I'd never heard it until fairly recently.) But your confusion is very common; it does seem like a weird thing to do when you first see it. So how can you get unconfused? It may help if you try replacing the variables (except the one you are solving for) with actual numbers, solve it that way (without doing any actual arithmetic) and then do the same thing using the variables. Here's an example: Let's solve P = 2(L+W) for W. I'll practice by solving this equation using numbers: 12 = 2(5 + W) To solve this, I have to undo everything that was done to W, one thing at a time. It was added to 5, then multiplied by 2; to undo those, in reverse order, we have to first divide by 2 and then subtract 5: 12/2 = 5 + W divided by 2 12/2 - 5 = W subtracted 5 Notice that I didn't bother to divide 12 by 2 to get 6, or subtract 5 to get 1, as I would normally do; I turned off the arithmetic part of my mind, and left only algebra running. But if I did the arithmetic, I would get the actual answer, 1. Now let's do exactly the same thing with the variables; P = 2(L + W) P/2 = L + W divided by 2 P/2 - L = W subtracted L That's the answer: W = P/2 - L Now, of course, if I plugged in the numbers P = 12 and L = 5, I would get W = 1. But having done all the algebra already, I could use different numbers and just use arithmetic (evaluate this expression) to find W again. That's what "literal equations" do for you; they predigest the algebra, leaving you needing only arithmetic to solve many similar problems. - 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/