subtracting a fraction containing a variable by another fraction without variables.

**Evan.Kimia** · September 28th 2009, 06:33 PM

I know this is easy, but i was wondering what the method is for doing this..

thank you!

**ramiee2010** · September 28th 2009, 06:50 PM

$= \ \frac{6(x+5)-8(x+3)}{6(x+3)} \times \frac{1}{(x-3)}$
$=\\ \frac{6x+30-8x-24}{6(x+3)(x-3)}$
$=\\ \frac{-2x+6}{6(x+3)(x-3)}$
$= \\ \frac{-2(x-3)}{6(x+3)(x-3)}$
$=\\ \frac{-1}{3(x+3)}$

**Evan.Kimia** · September 28th 2009, 07:18 PM

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

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