Math HELP!
posted by Sabrina on .
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?

I answered some of them
click on your name above and you will see your posts 
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?

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 
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

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.