Dividing Compound Fractions

Date: 11/16/1999 at 07:12:34
From: Rose Flynn
Subject: Compound Fractions Division

     3/y + 5/zy^2
     ------------
     5/2y - 4/y^2

Please help me understand how to solve such a problem. We are doing 
these in Algebra II and I don't understand the process. I need a 
simple step-by-step method. My textbook says I should find a creative 
form of one and multiply the minor denominator by it, but it only 
gives one example.

Date: 11/16/1999 at 12:45:55
From: Doctor Peterson
Subject: Re: Compound Fractions Division

Hi, Rose.

I think your expression looks like this:

       3     5
      --- + ----
       y    zy^2
     ------------
       5     4
      --- - ---
       2y   y^2

where the squares apply only to the y's. If I'm wrong, it probably 
doesn't make much difference.

Here are the steps for simplifying this sort of expression:

1. Find a common denominator for each set of fractions being added. 
That is, you want the common multiple of y and zy^2, and of 2y and 
y^2. Since y is a divisor of zy^2, you can use the latter as the 
common multiple. For 2y and y^2, you can use 2y^2.

2. Convert each fraction to the appropriate denominator. For example, 
you want to rewrite 3/y as ?/zy^2, so you have to multiply the 3 by 
zy. This is probably what they meant by multiplying by a creative form 
of 1; what we're doing is multiplying 3/y by zy/zy, whose value is 1 
so it doesn't change the value of the fraction.

3. Now add (or subtract) the fractions.

4. Now divide the two single fractions you have left, by multiplying 
the top fraction by the reciprocal of the bottom fraction.

That's it.

There is a shortcut you could take if you were very comfortable with 
this sort of fraction: find the LCM of ALL FOUR denominators, and 
multiply each fraction by that. But don't do that if such a big step 
scares you - it isn't necessary. The slow, plodding way works just 
fine.

Just to make sure you have it, I'll give an example for you to follow:

       2     3        2xy    3       2xy + 3
      --- + ----     ---- + ----     -------
       x    x^2y     x^2y   x^2y      x^2y      2xy + 3    3y^2
     ------------ = ------------- = --------- = ------- * -------
       4     5        4y     15      4y - 15     x^2y     4y - 15
      --- - ---      ---- - ----     -------
       3y   y^2      3y^2   3y^2      3y^2

Now we can cancel common factors of y and multiply:

     (2xy + 3)(3y)
     -------------
      x^2(4y - 15)

Finally, depending on the goal, you can distribute:

      6xy^2 + 9y
     -------------
     4x^2y - 15x^2

Let me know if you need any more help.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/