$(18B^2/(A^2-9B^2)) - (A/A+3B) + 2$ 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.


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*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?

$$\frac{18b^2}{a^2-9b^2}+2=\frac{2a^2}{a^2-9b^2}$$
As $a^2-9b^2=(a)^2-(3b)^2=(a+3b)(a-3b),$
$$\frac{18b^2}{a^2-9b^2}+2-\frac a{a+3b}=\frac{2a^2}{a^2-9b^2}-\frac a{a+3b}$$
$$=a\left(\frac{2a}{a^2-9b^2}-\frac1{a+3b}\right)$$
$$=a\cdot\frac{2a-(a-3b)}{(a+3b)(a-3b)}=a\cdot\frac1{a-3b}$$ assuming  $a+3b\ne0$
