Help me with some algebra

I need some help,

here are the problems:

i am subtracting these two fractions

(9b^2 - b + 4 / 3b - 2) - (2b^2 + b - 8 / 3b - 2)

The both look like fractions

Here is the second one i need help with

(11f / (2f + 3)(f - 4)) / (f + 6 / 2f - 3)

Some help would be great, thanks!

(9b^2 - b + 4 / 3b - 2) - (2b^2 + b - 8 / 3b - 2) For this first one the denominators are the same so you can just subtract the numerators.

so (9b^2-b+4-2b^2-b+8)/(3b-2)=

rearrange to see easier

(9b^2-2b^2-b-b+4+8)/(3b-2)=

(7b^2-2b+12)/(3b-2)<<< 1st answer

(11f / (2f + 3)(f - 4)) / (f + 6 / 2f - 3) In this one you have to multiply the first two and then do your division.

so (11f(f-4))/(2f+3)/(f+6/2f-3)=

[(11f^2-44f)/(2f+3)]/(f+6/2f-3)

now you two fractions, one divided by the other. So you multiply the first by the reciprocal of the other (the bottom one).

(11f^2-44f)(2f-3)/(2f+3)(f+6)=

distribute and simplify
(22f^3-88f^2-33f^2+132f)/
(2f^2+3f+12f+18)=

(22f^3-121f^2+132f)/(2f^2+15f+18)<<< 2nd answer

I hope this helps. E-mail if you have questions.

Problem A:
The answer is 7b^2+2b

Problem B:
The answer is

11f
___
(4f-3)(2f+3)(f-4)

I like using www.mathway.com to help solve my math problems.