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OpenStudy (anonymous):
@skullpatrol
OpenStudy (anonymous):
@e.cociuba
OpenStudy (anonymous):
@ganeshie8 @amistre64
OpenStudy (anonymous):
its quite simple really
5(x+2)/(x+2)x 2x/2(x-2)
x+2 and 2 cancel
5x/(x-2) is answer
OpenStudy (anonymous):
how does x+2 and 2 cancel?
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OpenStudy (e.cociuba):
(5x+10)/(x+2)*(2x)/(4x-10)
(5(x)+5(2))/(x+2)*(2x)/(4x-10)
(5(x+2))/(x+2)*(2x)/(4x-10)
(5 (x+2) )/( (x+2) )*(2x)/(4x-10) 5*(2x)/(4x-10)
5*(2x)/(2(2x)+2(-5))
5*(2x)/(2(2x-5))
5*(x)/(2x-5)
(5x)/(2x-5) is the answer.
OpenStudy (e.cociuba):
I did it the longer way, but @ArslanQureshi u can take this one :))
OpenStudy (anonymous):
no x+2 and x+2 in the first term and 2 and 2 in the second one and @e.cociuba you can have one from me for your hard work :)
OpenStudy (e.cociuba):
Haha thanks :)
OpenStudy (anonymous):
thnx guys i really apreciate it. i have another question
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OpenStudy (anonymous):
ask :)
OpenStudy (anonymous):
3n^2-n/n^2-1 divided by n^2/n+1
OpenStudy (anonymous):
3n^2-n/n^2-1 / n^2/n+1
n(3n-1)/n^2-1 x n+1/n^2
(3n-1)/(n+1)(n-1) x(n+1)/n---- n in the first term gets cancelled with n^2 leaving n
n(3n-1)/(n-1) is your answer ---- (n+1)s cancel
OpenStudy (anonymous):
so the answwer would be 3n?
OpenStudy (anonymous):
no the term in the last line is the answer n(3n-1)/(n-1)