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Algebra 7 Online
OpenStudy (anonymous):

divide. 3n^2-n/n^2-1 divided by n^2/n+1

OpenStudy (mertsj):

\[\frac{3n^2-n}{n^2-1}\div \frac{n^2}{n+1}\]

OpenStudy (mertsj):

First change it to a multiplication problem by inverting the second fraction.

OpenStudy (anonymous):

wat is inverting is thst like the reciprocal

OpenStudy (mertsj):

yes. Change it to its reciprocal.

OpenStudy (mertsj):

And then factor the numerator and denominator of the first fraction.

OpenStudy (anonymous):

then u get 3 divided by n+1 over n^2?

OpenStudy (mertsj):

\[\frac{n(3n-1)}{(n-1)(n+1)}\times \frac{n+1}{n^2}=\frac{3n-1}{n(n-1)}\]

OpenStudy (anonymous):

then u get 3 over n as the answer or just 3

OpenStudy (mertsj):

Did you read what I wrote?

OpenStudy (mertsj):

That fraction will not reduce any further.

OpenStudy (anonymous):

oh okay thank you

OpenStudy (mertsj):

yw

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