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```Date: 10/26/2002 at 18:54:32
From: Jay Jay
Subject: Algebra with variables

What is ax + bc = cx - df solving for x?

I do not understand how you can divide and add and subtract letters.
This whole entire problem stumps me!

Thank you - I really appreciate it.
```

```
Date: 10/26/2002 at 20:25:27
From: Doctor Ian
Subject: Re: Algebra with variables

Hi Jay Jay,

A variable is just a number whose value we don't happen to know yet.

Think of it this way. Suppose I have an envelope with some money in
it, but we don't know how much. We decide that we're going to split
the money evenly once the envelope is finally opened, but that's not
going to be for a while yet.

What can you say about how much you're going to get?  You can't tell
me the actual amount, but you could say

(the amount I'll get) = (the amount in the envelope) / 2

Does that make sense? It's sort of a promise to do the arithmetic
later. Of course, writing all those words out is inconvenient, so we
choose single letters to stand for them. For example, we might use A
to represent the amount you'll get, and E to represent the amount in
the envelope:

A = E / 2

So I'm not really dividing a _letter_ by anything. I'm saying that
once I know what _number_ E stands for, I'll divide _that_ by 2, and
the result will be what A stands for. Okay so far?

Now, this is nice if I expect to be told the value of E. But what if
I let _you_ open the envelope, take out half, and hand the rest to me?
How can I figure out how much was in the envelope to start with?

In this case, I know the value of A, and I want to know the value of
E.  That is, instead of something that looks like

A = E / 2

I want to have something that looks like

2 * A = E

These two equations say exactly the same thing. The only difference is
that in the first, I start with the value of E, do some operations,
and end up with the value of A; while in the other, I start with the
value of A, do some operations, and end up with the value of E.

We say that in the first case, the equation has been 'solved for A',
and in the second case, the equation has been 'solved for E'.  Once
again, it's the same equation in both cases! It's just presented in
two different ways.

So now let's look at an equation like

A = L * W

which is the formula for the area of a rectangle. (A represents the
area; L and W represent the length and width of the rectangle.)

Now, if we know the length and width, we're all set. We just plug
the values in and simplify to get the area. For example, if the
length is 6 inches and the width is 4 inches,

A = L * W

= 6 * 4

= 24 square inches

But what if we already know the area, and the length, but we don't
know the width?  We could still plug the numbers in:

24 = 6 * W

but now we have to monkey around with it to find the width. Or, we can
note that if two things are equal, they remain equal if we do the same
thing to both. We know that A and (L*W) are equal. If we divide them
both by L - regardless of what value L turns out to have - we'll end
up with two things that are still equal:

A   L * W
- = -----
L     L

Now the L's cancel on the right side, just as they would with
something like

2 * W
... = -----
2

to leave us with this:

A/L = W

This is just what we want. Now we can plug in the area and the length,
and the width pops right out:

W = 24/6

= 4

With equations like

ax + bc = cx - df

things can get a little more complicated, but it's exactly the same
idea:

ax + bc = cx - df

Subtract ax from both sides:

bc = cx - df - ax

bc + df = cx - ax

Distributive property:

bc + df = (c - a)x

Can you see what the final step would have to be, and why it works?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra

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