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### Manipulating a Formula Algebraically

Date: 06/15/2006 at 23:45:03
From: Debbie
Subject: What is this formula used for?

I have the formula M = C(1 + r) in my algebra text, but it doesn't
tell me what it is for.  It also asks me to solve for the r variable.

I know that the formula when solved for the r variable is:

M - C
r = -------
C

but I don't understand how to get all the way there.  This is as far
as I know:

M = C(1 + r)
-C  -C
----------------
M - C =  (1 + r)

Beyond this I haven't got a clue.  Where did the 1 go?  Where does the
extra C come from?  How do you get r alone?  Thanks for your help!

Date: 06/16/2006 at 10:17:55
From: Doctor Peterson
Subject: Re: What is this formula used for?

Hi, Debbie.

First, you don't need to even ask what a formula is for in this sort
of problem.  One of the most important things about algebra (or math
in general, really) is that it is abstract: that is, we can take all
sorts of real-world problems and turn them into math problems (such
as equations to solve), and once you've done that, it doesn't matter
at all where they came from.  The methods you use in algebra ignore
the meaning of the problem, and just look at the equation itself.

Many equations you'll work with in class don't come from anywhere at
all; you are just practicing techniques that you can use on problems
that do have some real-world meaning.  It's sort of like a medical
student practicing an operation on a dummy; he doesn't have to ask
about the patient's family or insurance!  But when he gets into real
medicine, all those things will matter.  In this case, the equation
MIGHT relate to interest on a bank account, with r being the interest
rate; but the same equation could come from other sorts of problems,
too--or none at all.

Now, to solve this equation for r, the important thing is to make
sure that at each step you are making a new equation that is still
true--that is, an equivalent equation.  We know that if we do the
same thing to each side, e.g. adding the same thing, the new equation
will be equivalent.  But often students don't pay close attention to
what they are doing, and they don't really make an equivalent
equation.

Look closely at what you did:

M = C(1 + r)
-C  -C
----------------
M - C =  (1 + r)

What you say you are doing is subtracting C from both sides.  But
that's not really what you did.  When you do the subtraction (without
simplifying anything), you actually get

M - C = C(1 + r) - C

Now you have to simplify, and you can't just cross off the C's!  If we
expand C(1 + r) using the distributive property, we get

M - C = C + Cr - C

and then combining like terms gives

M - C = Cr

That's not what you got!  Why?  Because you didn't actually subtract C
from the right side, but just crossed something off, thinking it was
the right thing to do.  This is a very common mistake!  You must
always think of the subtraction as making a change to the equation and
then simplifying, not just as canceling something.

We can continue, now--even though what you did is not the recommended
method, which I will get to in a minute.  Look at the equation we have
now:

M - C = Cr

What is our goal?  To get r by itself.  We're almost there; all that's
"wrong" with this is that r is multiplied by C, and we can get rid of
that by dividing by C.  So let's divide BOTH SIDES by C, again making
sure that's really what we do:

M - C   Cr
----- = --
C     C

Now we can simplify the right side; multiplying r by C and then
dividing by C undoes the multiplication and just leaves r:

M - C
----- = r
C

Now, there are two other methods that make the first step easier than
what we ended up doing.  One is to always simplify both sides of an
equation before we start solving:

M = C(1 + r)

M = C + Cr

(I distributed on the right side.)

Now we want to get r alone; on the right side it is FIRST being
multiplied by C (remember the order of operations?) and THEN we're
adding C to it.  To get it by itself, we have to undo both operations;
and we do that in reverse order.  (For example, in the morning I put
on my socks first, then my shoes; to undo that at night, I take off
my shoes first, then my socks.)  So we'll first undo the addition of
C, by subtracting C from both sides:

M   =   C + Cr
- C     - C
-----   --------
M - C =       Cr

What this really means is that we are subtracting C from both sides
like this:

M - C = C + Cr - C = Cr

where I simplified the right side by combining the like terms C and
-C.

Now we're right where we were in the first method, and just have to
divide by C to finish.

There's another way we could have done this, without simplifying
first; we'd look at the equation as given and see that in

M = C(1 + r)

we are FIRST adding 1 to r, and THEN multiplying by C.  We can undo
that by FIRST dividing both sides by C, and THEN subtracting 1 from
both sides.  The answer we get would look different, but would mean
the same thing.  If you wish, you may try doing that; but the method
I've shown is what is usually taught.  I mention it because it is
close to what you tried to do: you wanted to get rid of the C first;
but because it is being MULTIPLIED rather than added, you have to
DIVIDE by it rather than subtract, in order to eliminate it.  What you
really did was to subtract C from the left side and divide by C on
the right, which didn't give an equivalent equation.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
Middle School Equations

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