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Multiplying and Dividing Fractions


Date: 8/18/96 at 8:31:12
From: D.S.K.
Subject: Multiplying and Dividing Fractions

Hi Dr. Math! I am a first year algebra student and I was hoping you 
could help me with a few problems. 

The instructions are to perform the indicated operations and express 
the results in lowest terms.

Here is the first one:

x + 6        x
-----  *  -------
3xy       2x + 12


Is this the right answer?

                  2
 (x + 6)x        x + 6x        x(x + 6)      1   
-----------  = ------------ = ----------- = ---- 
3xy(2x + 12)     2            6xy(x + 6)     6y
               6x y + 36xy

_________________________________________________

Here is another problem:

  1         9x
-----  /  ------     
 2        x + 3
x - 9

Here is how I tried to do it:

  1       x + 3          1            x + 3   
------ * ------- = --------------- * ------- = 
 2         9x      (x - 3)(x + 3)      9x      
x - 9

   1(x + 3)              1
---------------   =  --------- .   Is this correct?
9x(x + 3)(x - 3)     9x(x - 3) 
  
_________________________________________________

How about this one?

 2         2      2    2
x - 2xy + y   /  x  - y
-----------     ---------
6xy                2  2
                 3x  y     

Is this how to do it?        

 2         2      2  2            2       2  2       
x - 2xy + y     3x  y      (x - y)      3x  y         
-----------  *  -------- = -------  * ------------- = 
    6xy          2    2      6xy      (x + y)(x - y)  
                x  - y

   2  2      2
3x  y (x - y)        xy(x - y)
----------------- =  --------- 
6xy(x - y)(x + y)    2(x + y)

_________________________________________________

I would really appreciate it if you could help me. 
Algebra can be kind of confusing sometimes.

D.S.K.


Date: 8/30/96 at 15:40:46
From: Doctor James
Subject: Re: Multiplying and Dividing Fractions

All your answers are right!

The first one you could have also done this way, which saves a little 
bit of work:

x + 6        x
-----  *  -------  = ( [x + 6]/3xy) * ( x/[2x + 12] )
 3xy      2x + 12
                   = ( [x + 6]/3xy) * ( x/2*[x + 6] )

                   = ( [x + 6]*x )/(2*3xy*[x + 6])

                   = x/6xy

                   = 1/6y,

which gives you the same answer, but saves you from having to multiply 
so much. However, your answer was still just as good.

See, algebra isn't that hard! They can make it bigger and more messy, 
but it doesn't get much harder than that.

-Doctor James,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
Middle School Algebra

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