1. ## Addition and subtraction with fractions

2. Originally Posted by klik11
$\frac{1}{3(a-1)} + \frac{a+2}{a(a+1)} - \frac{2(2a-1)}{3(a-1)(a+1)}$

common denominator is $3a(a-1)(a+1)$

$\frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}
$

now proceed ...

3. That exercise doesn't have it's solution on the book, so could you please tell me if I did it right?

http://img690.imageshack.us/img690/8339/algebra474.png

4. you have a mistake in the denominator.u didn't factorize well.

5. you should end up with ...

$\frac{2}{a(a+1)}$

6. Originally Posted by Raoh
well done "klik11".
Originally Posted by skeeter
you should end up with ...

$\frac{2}{a(a+1)}$

7. $\frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}$ $=\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $= \frac{6(a-1)}{3a(a^{2}-1)}$ $= \frac{2(a-1)}{a(a^{2}-1)}$.

8. I repeat ...

you should end up with $\frac{2}{a(a+1)}$

9. Originally Posted by Raoh
$\frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}$ $=\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $= \frac{6(a-1)}{3a(a^{2}-1)}$ $= \frac{2(a-1)}{a(a^{2}-1)}$.
which is $\frac{2}{a(a+1)}$

10. I didn't learn that part yet but okay
Thanks