Posted by Aria on Tuesday, August 2, 2011 at 1:58am.
I am stumped by these two LCM questions for some reason.
Find The LCM for
#1
2(R7) and 14(R7)
#2
(1+4r), (116r^2), and (14r)
It seems like #1 is already at the LCM
and #2 would be (1+4r)(14r)
Am I even vaguely correct?
Thanks for your help.

Algebra  MathMate, Tuesday, August 2, 2011 at 8:53am
As you probably know, if one number is a multiple of the other, the bigger number is the LCM of the two.
Post your exact answer to #1 to confirm if you wish.
For #2, you are more than vaguely correct, but you do not appear to have any justification of your response.
You can confirm your thoughts by dividing the LCM by each of the three values. If they all divide without remainder, and there is no common factor between the quotients, the answer is correct.
For example,
If 60 is the LCM of 5,6,15, then
60/5=12 has no remainder
60/6=10 has no remainder
60/15=5 has no remainder.
However,
5,10 and 12 have a common factor of 2, so 60 is not the LCM, but 60/2=30 is!
Feel free to post your justification of your answer along the lines above.