Simplify:
$$ \frac{18B^2}{A^2-9B^2} - \frac{A}{A+3B} + 2$$
If the notation doesn't work like I wrote it above it's; Simplify: $\frac{18B^2}{A^2-9B^2} - \frac{A}{A+3B} + 2.$
I made denominator common by expanding $A^2-9B^2=(A+3B)(A-3B)$ So the A after the minus should be still multiplied by $(A-3B)$ This gave me: $A^2-3AB.$
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$: \frac{6b-a}{3b-a} +2$
I made it smaller to $= 2+ \frac{2b-a}{b-a}$
None of this seems correct. I first added 2 to the end. I then tried to add $\frac42$ earlier. That made my final answer: $\frac{6B-A+4}{3B-A+2} = \frac{2B-A+4}{B-A+2}.$ I also tried times $\frac42.$ 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.