Basic fraction question "rewrite as single fraction"

• Jan 17th 2010, 01:21 PM
Wolvenmoon
Basic fraction question "rewrite as single fraction"
So I'm working calc 1 when I come to a problem with an odd fraction rule I just don't get.

The issue is this, I have (2/x) - (2/3 ) all over (x-3 ), one of the steps is to rewrite (2/x) - (2/3) as a single fraction, which turns out to be (6-2x)/3x

How did they do this?

( the full problem is: "The limit as X approaches 3 is "f(x)-f(3) all over (x-3)", f(x) = 2/x , and it works out to -2/9, but my hangup is in rewriting this fraction )
• Jan 17th 2010, 01:27 PM
General
Quote:

Originally Posted by Wolvenmoon
The issue is this, I have (2/x) - (2/3 ) all over (x-3 ), one of the steps is to rewrite (2/x) - (2/3) as a single fraction, which turns out to be (6-2x)/3x

How did they do this?

By using this wonderful formula ! :
$\frac{a}{b} - \frac{c}{d} = \frac{ ad - bc }{bd}$
• Jan 17th 2010, 01:31 PM
$\frac{2}{x}-\frac{2}{3}=\frac{3}{3}\frac{2}{x}-\frac{x}{x}\frac{2}{3}$