Addition and subtraction with fractions

**klik11** · November 8th 2009, 09:43 AM

What am I doing wrong?

http://img26.imageshack.us/img26/8339/algebra474.png

**skeeter** · November 8th 2009, 10:04 AM

Originally Posted by klik11

What am I doing wrong?

http://img26.imageshack.us/img26/8339/algebra474.png

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

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

$\frac{a(a+1)}{3a(a-1)(a+1)} + \frac{3(a-1)(a+2)}{3a(a-1)(a+1)} - \frac{2a(2a-1)}{3a(a-1)(a+1)}<br />$

now proceed ...

**klik11** · November 8th 2009, 10:35 AM

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

**Raoh** · November 8th 2009, 11:01 AM

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

**skeeter** · November 8th 2009, 11:04 AM

you should end up with ...

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

**klik11** · November 8th 2009, 11:18 AM

Originally Posted by Raoh

well done "klik11".

Originally Posted by skeeter

you should end up with ...

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

**Raoh** · November 8th 2009, 11:29 AM

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

**skeeter** · November 8th 2009, 11:29 AM

I repeat ...

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

**Raoh** · November 8th 2009, 11:31 AM

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

**klik11** · November 8th 2009, 12:27 PM

I didn't learn that part yet but okay

Thanks

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