The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

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 = -------

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 

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 

  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

And that's our answer!

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 

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 
Associated Topics:
Middle School Equations

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.