Question

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

Answer 1

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

Answer 2

factorising here, is the "key" to your solution.

Answer 3

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

is this right

Answer 4

YOu're on the right track. Now flip the second fraction upside down so that you can change the division sign to a multiplication sign.

Answer 5

You already flipped it.

Answer 6

sorry I was suppose to put multiply instead of divide

Answer 7

Changed the sign. Okay.

Answer 8

Then take away those common multiples.

Answer 9

Wait I don't think you can take any.

Answer 10

You sure you wrote down the correct equation. Because you can't take away any common values.

Answer 11

Ya that is what they gave me

Answer 12

Was your question: (3n^2-n) / (n^2-1) / (n+1)/n^2 meant to be this:(3n^2-n) / (n^2-1) / n^2/n+1?

Answer 13

look at the last fraction, that does seem dodgy to me. Could you double check your fractions to make sure each term is in the correct position.

Answer 14

\[\frac{ n ^{2} }{ n+1 }\]
this is the original

Answer 15

Ahah, I want to say told you so, but I won't. So now you flip that upside down. Now you can get rid of the common factors diagonally across each fraction.

Answer 16

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

Answer 17

That's what you would end up with, correct?

Answer 18

sorry I wrote it down and then flipped it back around.

Answer 19

No worries.And then you just combine the result to leave it in one fraction.

Answer 20

\[\frac{ 3n-1 }{ n-1 }\times \frac{ 1 }{ n }=?\]

Answer 21

What you fill in the question mark, is your answer.

Answer 22

\[\frac{ 3n-1 }{ n(n-1) }\]

Answer 23

You nailed it right into the head. Well Done. Excellent work.

Answer 24

Thank you for walking me through it.

Answer 25

No worries, anytime.