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divide. 3n^2-n/n^2-1 divided by n^2/n+1
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\[\frac{3n^2-n}{n^2-1}\div \frac{n^2}{n+1}\]
First change it to a multiplication problem by inverting the second fraction.
wat is inverting is thst like the reciprocal
yes. Change it to its reciprocal.
And then factor the numerator and denominator of the first fraction.
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then u get 3 divided by n+1 over n^2?
\[\frac{n(3n-1)}{(n-1)(n+1)}\times \frac{n+1}{n^2}=\frac{3n-1}{n(n-1)}\]
then u get 3 over n as the answer or just 3
Did you read what I wrote?
That fraction will not reduce any further.
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oh okay thank you
yw
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