What am I doing wrong? (Headbang)

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

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

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- Nov 8th 2009, 09:43 AMklik11Addition and subtraction with fractions
What am I doing wrong? (Headbang)

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

http://img26.imageshack.us/img26/8339/algebra474.png - Nov 8th 2009, 10:04 AMskeeter
- Nov 8th 2009, 10:35 AMklik11
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 - Nov 8th 2009, 11:01 AMRaoh
you have a mistake in the denominator.u didn't factorize well.

- Nov 8th 2009, 11:04 AMskeeter
you should end up with ...

$\displaystyle \frac{2}{a(a+1)}$ - Nov 8th 2009, 11:18 AMklik11
- Nov 8th 2009, 11:29 AMRaoh
$\displaystyle \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)}$$\displaystyle =\frac{a(a+1)+3(a-1)(a+2)-2a(2a-1)}{3a(a-1)(a+1)}$ $\displaystyle = \frac{6(a-1)}{3a(a^{2}-1)}$ $\displaystyle = \frac{2(a-1)}{a(a^{2}-1)}$.

- Nov 8th 2009, 11:29 AMskeeter
I repeat ...

you should end up with $\displaystyle \frac{2}{a(a+1)}$ - Nov 8th 2009, 11:31 AMRaoh
- Nov 8th 2009, 12:27 PMklik11
I didn't learn that part yet but okay :)

Thanks