I know this is easy, but i was wondering what the method is for doing this..http://img143.imageshack.us/img143/6223/capturefi.jpg

thank you!

- Sep 28th 2009, 06:33 PMEvan.Kimiasubtracting a fraction containing a variable by another fraction without variables.
I know this is easy, but i was wondering what the method is for doing this..http://img143.imageshack.us/img143/6223/capturefi.jpg

thank you! - Sep 28th 2009, 06:50 PMramiee2010
$\displaystyle = \ \frac{6(x+5)-8(x+3)}{6(x+3)} \times \frac{1}{(x-3)}$

$\displaystyle =\\ \frac{6x+30-8x-24}{6(x+3)(x-3)}$

$\displaystyle =\\ \frac{-2x+6}{6(x+3)(x-3)}$

$\displaystyle = \\ \frac{-2(x-3)}{6(x+3)(x-3)}$

$\displaystyle =\\ \frac{-1}{3(x+3)}$ - Sep 28th 2009, 07:18 PMEvan.Kimia
Thank you, but i meant from the middle part of the eq. where it has ((x+5)/(x+3))-(8/6).

Im not sure what method they used to subtract the 2. ~Evan