Simplifying Fractional Equations

Date: 4/4/96 at 16:52:37
From:  RDIANNE GREEN
Subject: Need help please

Dear Doctor Math,

I am home schooled, and I have been unable to answer the following 
algebra questions. I am hoping you can help give me insight into 
these problems:  

I am supposed to perform the indicated operations and express 
results in lowest terms.

1.  x^2 - 2xy + y^2          3x^2y^2
   __________________   *  ____________   =   ?
   
          6xy               x^2 - y^2

2.    5           4x         2 - x
   ________  -  _______  -  ________  =  ?

    x - 1        x - 1       x - 1

Thanks a lot in advance for your help.

Hope to get your response soon,

jodi

Date: 4/5/96 at 11:24:1
From: Doctor Aaron
Subject: Re: Need help please

Hi Jodi,

Working with fractions of algebraic expressions is just like 
working with fractions of integers.  One thing we have to do is 
find a common denominator.  Since you were asked to express the 
result in lowest terms, we also have to worry about that.

In problem 1 you have a product of two expressions.  If it were

5   6
- * - 
3  15

you'd want to do some canceling, but with algebraic expressions
it's a little harder to see what cancels with what and what 
will be left over.

With algebraic expressions we just have to define what we mean by 
a factor.

Well 6xy isn't too bad - it's just 6*x*y so the factors are 6, x 
and y.

Now we have to figure out x^2-y^2.  Well, we can rewrite this as
x^2 - xy + xy - y^2, which may seem silly at first, but it will be 
useful.  Then we can factor out an x from the first two pieces to 
rewrite it again as:

x*(x-y) + xy - y^2

We can do the same for y and the last two pieces to get
x*(x-y) + y*(x-y), which we can rewrite as:

     (x+y)*(x-y). 

Then the factors, or multiples, are (x+y) and (x-y) 

This is the process known as factoring.  We have broken up a
sum of variables to the second power to a product of sums of 
variables to the first power. 

Try to work through x^2 - 2xy - y^2 by yourself.  You should get 
(x-y)*(x-y).  

Once you believe that x^2-2xy-y^2 = (x-y)*(x-y), we can rewrite 
your equation as: 

        (x-y)*(x-y)      3*x^2*y^2
         ----------  *  ----------- 
           6*x*y        (x+y)*(x-y)

we can cancel an x-y to get

           (x-y)   3*x^2*y^2
           ----- * ---------  
           6*x*y      (x+y)  

but we can still simplify because we can write x^2 as x*x and y^2 
as y*y and then cancel to get:

(x-y)    3*x*y                                      (x-y)*x*y   
----- *  ----  but 3*2 = 6 so we can rewrite again: ---------
 6       (x+y)                                       2*(x+y)

The second problem is a little different.  Instead of a product, 
you have to calculate a difference.  Luckily you already have a 
common denominator, x-1.  You can multiply both sides of the 
equation by x-1 just as you would multiply by 3 if you saw:

          4   +    5   -   x     22  
          -        -       -  =  -
          3        1       3     3 

After you multiply both sides by x-1, the denominators will cancel 
and you'll be left with 5 - 4x = 2 - x.

 Good luck.

-Doctor Aaron,  The Math Forum