Hi Everyone! I would like you to check my work.

BTW my book is Precalculus, 3rd Edition by Bittinger, Beecher, Ellenbogen, and Penna

The directions are to add/subtract/divide and simplify if possible.

#38: $\displaystyle \frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{2b\,-\,3a}$

Get like denominators: $\displaystyle \;\frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

$\displaystyle \;=\frac{-3a}{-3a\,+\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

Simplify: $\displaystyle \;\frac{\,-\,5a}{-3a\,+\,2b}$

Would it be better to put my answer as: $\displaystyle \;\frac{5a}{3a\,-\,2b}$ ?

# 39: $\displaystyle \frac{9x\,+\,2}{3x^2\,-\,2x\,-\,8}\,+\,\frac{7}{3x^2\,+\,x\,-\,4}$

Factor the denominators: $\displaystyle \frac{9x\,+\,2}{(3x\,+\,4)(x\,-\,2)}\,+\,\frac{7}{(3x\,+\,4)(x\,-\,1)}$

Get a LCM of the denominators: $\displaystyle \frac{(9x\,+\,2)(x\,-\,1)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}\,+\,\frac{7x\,-\,14}{(3x\,+\,4)(x\,-\,1)(x\,+\,2)}$

$\displaystyle \;=\,\frac{9x^2\,-\,7x\,-\,2\,+\,7x\,-\,14}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Simplify: $\displaystyle \frac{(3x\,-\,4)(3x\,+\,4)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Which simplifies to:

$\displaystyle \frac{3x\,-\,4}{(x\,-\,2)(x\,-\,1)}$

My book agrees so I'm pretty sure.

#44: $\displaystyle \frac{6}{x\,+\,3}\,-\,\frac{x\,+\,4}{9\,-\,x^2}\,+\,\frac{2x\,-\,3}{9\,-\,6x\,+\,x^2}$

My answer is: $\displaystyle \frac{-8x^2\,+\,32x\,-\,41}{-(x\,-\,3)^2(x\,+\,3)}$

# 62: $\displaystyle \frac{1}{x^2}\,-\,\frac{1}{y^2}\div\frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2}$

Get the like denominators: $\displaystyle \frac{y^2\,-\,x^2}{x^2y^2}\,\div\,\frac{y^2\,-\,2x\,+\,x^2}{x^2y^2}$

$\displaystyle \;=\,\frac{y^2\,-\,x^2}{x^2y^2}\cdot\,\frac{x^2y^2}{y^2\,-\,2x\,+\,x^2}$

Which just simplifies to: $\displaystyle \frac{y^2\,-\,x^2}{y^2\,-\,2x\,+\,x^2}$

Correct?

Thank you all for your help!