How do I divide algeabraic fractions?

Question

Okay, so here is the problem I am stuck on:

3 3n-9
------- divided by --------- What's the answer?
n^2-9 n+3

Then there's one I want to double check upon. Which is:
c d
-- divided by -----
2d 8c

for an answer I got: 4c^2
--------
d^2
is that right?

Answer 1

(3/n^2-9) / (3n-9/n+3)

First: rule - you can't divide fractions > multiply the 1st fraction by the reciprocal of the 2nd fraction...

(3/n^2-9) * (3n-9/n+3)

Sec: factor the denominator in the 1st fraction & factor the numerator in the 2nd fraction...

[3/(n+3)(n-3)] * [3(n-3)/n+3)]

Third: cross cancel "like" terms...

3 * 3
= 9

Answer 2

The first problem as shown is the answer. The top could be reduced to 3*(11n-3) but that doesn't help much because the bottom can't be factored. The general format would be (n+/-#) * (n+/-#), but it can not be factored. For instance if the bottom term were n^2- 4n+3 then it could be factored thus (n-1)*(n-3). Once you have the polynomial factored you can cancel like terms. See below.
The answer to the second problem is 1/16. The variables "c" and "d" in the top can be divided by the same variable in the bottom. Therefore c/c is 1 and d/d is 1. Thus the result is 1*1 / 2*8 = 1/16.

Answer 3

i'm not sure about the first one..but that second one is 1/28 i believe. if you factor out a cd on the top and bottom and then cancel them.