$(18B^2/(A^2-9B^2)) - (A/A+3B) + 2$ Simplify:
$$ \frac{18B²}{A²-9B²} - \frac{A}{A+3B} + 2$$
If the notation doesn't work like I wrote it above it's; Simplify: 18B^2/A^2-9B^2 - A/A+3B + 2.


*

*I made denominator common by expanding A²-9B²: (A+3B)(A-3B) 
So the A after the minus should be still multiplied by (A-3B)
This gave me: A²-3AB.

*Then I put common factors together but left out the +2. This is where I probably go wrong. Without the 2 I expanded:(18B^2-a^2+3ab) to (6B-A)(3B+A). The denominator was (3B-A)(3B+A). I cancelled out (3B+A) on both sides. I was left with: 6b-a/3b-a +2  

*I made it smaller to = 2+ 2b-a/b-a
None of this seems correct. I first added 2 to the end. I then tried to add 4/2 earlier. That made my final answer: 6B-A+4/3B-A+2 = 2B-A+4/B-A+2. I also tried times 4/2. That also did not give me the correct answer and didn't seem logical. Perhaps I shouldn't have cancelled out? But I wouldn't know why. What am I doing wrong?

You made a small mistake in step 2 and a bigger mistake in step 3.  The small mistake is that the denominator is not $(3B-A)(3B+A)$, but $(A-3B)(A+3B)$ -- in other words, you're off by a minus sign.  So at the end of step 2 you should have
$${6b-a\over a-3b}+2$$
The bigger mistake is thinking you can cancel the $3$ with the $6$, turning this into ${2b-a\over a-b}+2$.  Instead you should change the $2$ into a fraction the denominator $a-3b$:
$${6b-a\over a-3b}+2={6b-a\over a-3b}+{2(a-3b)\over a-3b}={6b-a+2a-6b\over a-3b}={a\over a-3b}$$
BTW, in calling the mistakes "small" and "bigger," I just mean that the sign error in step 2 struck me as very likely an oversight -- it's easy to get things turned around when factoring expressions -- but the cancellation error in step 3 struck me as possibly a more serious conceptional misunderstanding.
