Question

(3n^2-n)/(n^2-1)divided(n^2)/(n+1)

Answer 1

factor and cancel i believe is what you need

Answer 2

start with
\[\frac{3n^2-n}{n^2-1}\times \frac{n+1}{n^2}\]

Answer 3

then factor as
\[\frac{n(3n-1)(n+1)}{(n+1)(n-1)n^2}\]

Answer 4

then cancel some stuff
\[\frac{\cancel{n}(3n-1)\cancel{(n+1)}}{\cancel{(n+1)}(n-1)n\cancel{n}}\]

Answer 5

it was divided by, i flipped and multiplied

Answer 6

just like with numbers
\[\frac{4}{5}\div \frac{8}{9}=\frac{4}{5}\times \frac{9}{8}\]

Answer 7

Oh ok so the answer is (3n-1)/(n-1)?