Simplifying Exponents in Fractions From my Algebra 2 class. Not homework.
$$4x^3/2x^5y^2$$
Divide the bases and subtract the exponents:
$$2x^{-2}y^2$$
Get rid of negative exponent by division:
$$2y^2/x^2$$
Then the answer should be:
$$2x^2y^2$$
Is this correct?
 A: Here's how I think about the exponent rules. It may be helpful to you. Reduce the numerical part first and consider 
$$
\frac{2x^3}{x^5y^2}.
$$
I notice that both have common factors of $x$. How many copies of $x$ are there? There are 3 on the top and 5 on the bottom, so I could think of the fraction as
$$
\frac{2xxx}{xxxxxy^2}.
$$
(Of course I would never actually write that down, but it's useful to keep in mind.) Now, if I went by and canceled factors of $x$ one-by-one, I would eventually be left with
$$
\frac{2}{xxy^2},
$$
or better yet
$$
\frac{2}{x^2y^2}.
$$
The advantage of thinking in this way is I don't memorize unmotivated rules about when to add or subtract exponents (though, if you reflect for a moment, you'll see that this method is the same as the exponent rules you've learned) and avoids getting negative exponents unnecessarily.
A: One of the problems with the "slant bar" notation is that it is not clear what fraction you have to begin with.
Does "$2x^2/x^5y^2$" represent
$$\frac{2x^2}{x^5y^2},$$
or does it represent
$$\frac{2x^2}{x^5}\cdot y^2\quad?$$
Normally, it would be interpreted as the first; but you seem to be interpreting it as the second. That would happen if, in the board or handwriting, there was a prominent and clear space, something like
$$2x^2/x^5\ \ y^2,$$
which got lost along the way.
Nonetheless, because of this possibility of confusion, I strongly advice all my students to abandon the "slant bar" notation in mathematical formulas. 
If your original problem was
$$\frac{2x^2}{x^5y^2}$$
then your first step is wrong, because the $y^2$ is in the denominator. The final answer should be as Austin Mohr gives it. 
If your original problem was
$$\frac{2x^2}{x^5}\cdot y^2,$$
then everything you did was right up to the final step; you should not have gone from $\displaystyle\frac{2y^2}{x^2}$ to $\displaystyle 2y^2x^2$.
