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### Solving Literal Equations

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Date: 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
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

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/
```
Associated Topics:
Middle School Algebra

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