# Addition and subtraction with fractions

• November 8th 2009, 09:43 AM
klik11
• November 8th 2009, 10:04 AM
skeeter
Quote:
$\frac{1}{3(a-1)} + \frac{a+2}{a(a+1)} - \frac{2(2a-1)}{3(a-1)(a+1)}$

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 ...
• November 8th 2009, 10:35 AM
klik11
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
http://img690.imageshack.us/img690/8339/algebra474.png
• November 8th 2009, 11:01 AM
Raoh
you have a mistake in the denominator.u didn't factorize well.
• November 8th 2009, 11:04 AM
skeeter
you should end up with ...

$\frac{2}{a(a+1)}$
• November 8th 2009, 11:18 AM
klik11
Quote:

Originally Posted by Raoh
well done "klik11".(Happy)

Quote:

Originally Posted by skeeter
you should end up with ...

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

(Thinking)
• November 8th 2009, 11:29 AM
Raoh
$\frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}$ $=\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $= \frac{6(a-1)}{3a(a^{2}-1)}$ $= \frac{2(a-1)}{a(a^{2}-1)}$.
• November 8th 2009, 11:29 AM
skeeter
I repeat ...

you should end up with $\frac{2}{a(a+1)}$
• November 8th 2009, 11:31 AM
Raoh
Quote:

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 $\frac{2}{a(a+1)}$
• November 8th 2009, 12:27 PM
klik11
I didn't learn that part yet but okay :)
Thanks