Posted by Sabrina on Friday, September 24, 2010 at 6:07pm.
I posted a few questions and then went back and posted how I thought they were supposed to be answered. Could someone please locate my earlier questions and tell me if I am on the right track?

Math HELP!  Reiny, Friday, September 24, 2010 at 6:25pm
I answered some of them
click on your name above and you will see your posts

Math HELP!  Sabrina, Friday, September 24, 2010 at 6:41pm
Thank you Reiny so much for you help. Can you tell me if I am on the right track with this problem: 4c^2/4c^28c+4 x 4c4/2c When solving this am I supposed to find the gcf or lcm to solve> would the lcm be 2 and the gcf be 16? I don't know if I am even going down the right road here?

Math HELP!  Reiny, Friday, September 24, 2010 at 7:11pm
Use brackets to show when the denominator ends
I think you meant
4c^2/(4c^28c+4)( 4c4)/2c , which is
= 4c^2/(4(c^2  2c + 1)(4)(c1)/(2c)
= 4c^2/(4(c 1)^2(4)(c1)/(2c)
now things cancel rather nicely
final result
2c/(c1) , c ≠ 1,0

Math HELP!  Sabrina, Friday, September 24, 2010 at 7:35pm
THANKS!!!!!! Just one question, when I follow these steps and get to the result is it both 2c/(c1), c‚ 10 or am I reading my problem wrong and the answer is just c‚ 10

Math HELP!  Reiny, Friday, September 24, 2010 at 9:38pm
The final answer is
2c/(c1)
the part after it is the restriction.
you are reading it wrong, it does not have 10 in it.
it says, c cannot be equal to 1 or c cannot be equal to 0
Since we cannot divide by zero, those values would make the denominators we canceled equal to zero.
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