Question 26470
<pre><font face = "lucida console" size=4><b>You either mean this:

9y<sup>4</sup>     (x<sup>-1</sup>)<sup>-2</sup>
———— + —————————
x<sup>-2</sup>       y<sup>2</sup>

or this:

9y<sup>4</sup> + (x<sup>-1</sup>)<sup>-2</sup>
—————————————
   x<sup>-2</sup>y<sup>2</sup> 

I'll do both:

In either case, you must learn the rule for getting 
rid of negative exponents.

If a numerator (respectively, denominator) either is 
or has a factor which is an exponential with a 
negative exponent, then move the base and exponent 
to the denominator (respectively, to the numerator)
and change the sign of the exponent to positive.

If it's this way:

9y<sup>4</sup>     (x<sup>-1</sup>)<sup>-2</sup>
———— + —————————
x<sup>-2</sup>       y<sup>2</sup>

  
In the first fraction, move the x<sup>-2</sup> from the 
denominator to the numerator and change the sign of
the -2 exponent to +2 and write x<sup>2</sup> in the numerator, 
and put a 1 in the denominator:

9y<sup>4</sup>x<sup>2</sup>     (x<sup>-1</sup>)<sup>-2</sup>
————— + —————————
  1         y<sup>2</sup>

In the second fraction multiply the outer exponent 
-2 by the inner exponent -1, getting +2 so it ends
up having a positive exponent, and there is no 
negative exponent to get rid of:

In the first fraction, move the x<sup>-2</sup> from the 
denominator to the numerator and change the sign of 
the -2 exponent to +2 and write x<sup>2</sup> in the numerator, 
and put a 1 in the denominator:

9y<sup>4</sup>x<sup>2</sup>    x<sup>2</sup>
————— + ———
  1      y<sup>2</sup>

Now the LCD is y<sup>2</sup>, so to make the first fraction 
have this LCD for its denominator, multiply top 
and bottom by y<sup>2</sup>

9y<sup>4</sup>x<sup>2</sup>·y<sup>2</sup>    x<sup>2</sup>
———————— + ———
  1·y<sup>2</sup>      y<sup>2</sup> 

Add the exponents of y in the numerator of the 
first fraction, and eliminate the "1·" in the bottom

9y<sup>6</sup>x<sup>2</sup>    x<sup>2</sup>
————— + ———
  y<sup>2</sup>     y<sup>2</sup>

Add the numerators and place over the common 
denominator

9y<sup>6</sup>x<sup>2</sup> + x<sup>2</sup>
——————————
     y<sup>2</sup>     

You may not do any canceling here.  You can, however, 
factor out an x<sup>2</sup> from the numerator, if you like:

x<sup>2</sup>(9y<sup>6</sup> + 1)
——————————
    y<sup>2</sup>

=====================================
 
If you meant this:

9y<sup>4</sup> + (x<sup>-1</sup>)<sup>-2</sup>
—————————————
   x<sup>-2</sup>y<sup>2</sup> 

Then In the second term on top, multiply the outer
exponent -2 by the inner exponent -1 getting +2 so 
it ends up having a positive exponent, and there is
no negative exponent to get rid of:

9y<sup>4</sup> + x<sup>2</sup>
————————
  x<sup>-2</sup>y<sup>2</sup>

Now x<sup>-2</sup> is a factor of the denominator, so move it 
from denominator to numerator by changing the sign 
of the -2 exponent and writing it in the top as x<sup>2</sup>.
However it must be a FACTOR of the numerator, so the
two terms in the numerator must by enclosed in 
parentheses:

x<sup>2</sup>(9y<sup>4</sup> + x<sup>2</sup>)
————————————
    y<sup>2</sup>

This is as far as you can go.

Edwin
AnlytcPhil@aol.com</pre>