SOLUTION: simplify: 16x^5y^4 over 32xy^7

 16x5y4 
 32xy7

Give the exponent 1 to the letter x in the bottom

 16x5y4 
 32x1y7

Cancel the 16 and the 32, putting 2
below the 32 since 32÷16 = 2, and
putting 1 above the 16, since 16÷16 = 1

     1
 16x5y4 
 32x1y7
     2

So now you have

 1x5y4 
 2x1y7

Now use the rule for subtracting exponents:

When two exponential with the same base
appear as a factor of the numerator and
the denominator:

Subtract the smaller exponent from the larger
exponent and place 

(a) the resulting exponent with the common base
in the numerator if that's where the larger 
exponent was previously and eliminate it from 
the denominator.   Or,

(b) the resulting exponent with the common base
in the denominator if that's where the larger 
exponent was previously and eliminate it from 
the numerator. 

So here we have a case of each

 1x5y4 
 2x1y7

We subtract the exponent of x1 from the exponent 
of x5, and get x4, and we place x4 in the numerator
because the larger exponent 5 was in the numerator,
and we eliminate x from the denominator:

 1x4y4 
 2y7

We subtract the exponent of y4 from the exponent 
of y7, and get y3, and we place y3 in the denominator
because the larger exponent 7 was in the denominator,
and we eliminate y from the numerator:


 1x4 
 2y3

Now we can erase the 1 coefficient in the top

  x4 
 2y3

Edwin