Algebra: Common Denominators

Date: 5/14/96 at 15:43:56
From:  RDIANNE   GREEN
Subject: Algebra help needed, please :)

Hi. I am having problems working out the following two problems 
because I don't know how to get all of them to have equal 
denominators.  

1.    2             a  +  1           4
   -------   +    ----------   =    -----
    a - 3          a^2 - 9          a + 3


2.   x            1
   -----   =    -----
   x + 2          x

Please help.  Thanks in advance.  :)

Date: 5/14/96 at 21:12:59
From: Doctor Jodi
Subject: Re: Algebra help needed, please :)

Hi there! You may remember that one common denominator (though not 
always the lowest) is found by multiplying the two (or more) 
denominators together.

For example, for the fractions 2/7 and 1/3, you can multiply 7 * 3 to 
get 21, one common denominator.

For your first problem, a^2 - 9 will be the lowest common denominator.  
[I know this because of the "difference of squares" factoring rule.  
Since a^2 - 9 is the difference of squares, we can factor it as 
(a - 3)(a+3)]

So for the first problem, we will need to multiply fractions with the 
denominator of a-3 by a+3, (which equals 1) and fractions with the  
                      ---
                      a+3

denominator of a+3 by a-3 (also equal to 1).  Does this make sense?
                      ---
                      a-3 

2       a + 3  =  2(a+3)    = 2a+6
--   * ------     ------      -----
a-3     a + 3     a^2 -9      a^2 -9


We can leave a+1     as it is for the moment.
             ---
             a^2 -9

We will need to multiply 

4          a-3                4 (a-3)         4a -12
-     *    ---     to get     -------    =    ------
a+3        a-3                 a^2 -9         a^2 -9


So now the problem reads


 2a +6 +a+1  = 4a -12
-----------    ------
a^2 -9         a^2 -9

Does that look better?

You can now multiply through by a^2 -9 to remove the denominators if 
you like and continue solving this like any other algebra problem.
Are you comfortable with the rest of the problem?  If not, write us 
back and we'll give you a hand.

Here's just a clue for the second problem.  Again, if you want more 
help, let us know.  For the second problem, the common denominator is 
x(x+2) (aka x^2 + 2x)

Good luck and thanks for your question!
-Doctor Jodi,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/