Not quite. What did you do with the (2) and the (3) to get its LCM? That's what you have to do with the (2) and the (x+2) in this situation.
Aha. You're missing a very important step. Let's say you're adding the fractions
$\displaystyle \dfrac{1}{2}+\dfrac{1}{3}.$
We've already determined that the LCM of the denominators, which is the same thing as the common denominator, is 6. What would your next step be in this addition problem?
No, no, no. You've found the common denominator. Now you have to get the common denominator. You do that by multiplying fractions by fancy ways of writing 1 in order to obtain the common denominator in each denominator. Let me illustrate:
$\displaystyle \displaystyle \frac{1}{2}+\frac{1}{3}=\frac{1}{2}\left(\frac{3}{ 3}\right)+\frac{1}{3}\left(\frac{2}{2}\right).$
So far, you would agree I haven't changed anything, right? I've multiplied each fraction by something that's equal to 1 (the fractions that are in parentheses). But I've chosen those numbers carefully so that when I multiply out the denominators, I'll get the common denominator that I need. Thus, the next step is to multiply out the fractions, which you can do the way you want to do it:
$\displaystyle \displaystyle\frac{1}{2}\left(\frac{3}{3}\right)+\ frac{1}{3}\left(\frac{2}{2}\right)=\frac{3}{6}+\fr ac{2}{6}.$
Do you see how this is done? And do you know what the next steps are?
Ok. So, picking back up from where we left off at post # 21, you're trying to add the two fractions
$\displaystyle \dfrac{x-1}{2}-\dfrac{4}{x+2}.$
You've determined that the least common denominator is $\displaystyle 2(x+2).$ Now you have to get the common denominator. What do you get?