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\[3n^2-n/n^2-1 \div n^2/n+1\] that make more sense? where the "/" its one piece of the problem over the other, please someone give me the typed out version of the answer without it being numbers divided over other numbers just right it out has one line :)
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\[\frac{3n^2-n}{n^2-1}\div \frac{n^2}{n+1}\] first step is to flip the second one and get \[\frac{3n^2+n}{n^2-1}\times \frac{n+1}{n^2}\] second step is to factor third step is to cancel
\[\frac{n(3n-1)(n+1)}{(n+1)(n-1)n^2}\]
cancel on factor of \(n\) and get rid of both \(x+1\) and you are done
thanks!
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