Could someone please show me in detail how you would go about finding the LCD for a problem such as this, with multiple variables and terms in the denominator.
Many Thanks
The denominators are very nearly factored. You should be able to do it by inspection.
First) 2, (x-1), (x+1)
Second) 2, 3, (x-1), (x-1)
Just read off the LCD by meeting the needs of all denominators.
2, (x-1), (x+1), 3, (x-1) again. ==> 2*3*(x+1)*(x-1)^2
Look at my response again. There is a reason why I wrote them in that order. I'll do it again witht he thought process outloud.
Just write down the first denominator
2, (x-1), (x+1)
Now the second, one piece at a time.
2 we have already. No need to add another.
3 is new. Add that.
(x-1) we have already. No need to add another.
(x-1) again. Whoops, there are two. We DO need another.
Second) 2, 3, (x-1), (x-1)
We did not DROP anything. We just met the needs of everyone. If you include one (x-1) for the first and two more from the second, that's a total of three and NO denominator has that many. We want the LEAST possible. All three is excessive.