Question

Please help!!!! I don't understand how to do this!!!

Answer 1

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

Answer 2

@tkhunny ?

Answer 3

Factor Factor Factor

Answer 4

Well how do I do that?

Answer 5

n^2 - 1 = (n+1)(n-1)

Factoring. You must have done it, before.

Answer 6

you could switch the second fraction and them multiply both fractions

Answer 7

that's what you need to do first ^ then you factor

Answer 8

Ok. Is this how it is? \[\frac{ 3(n+1)(n-1) }{ (n+1)(n-1) }\]

Answer 9

@johnweldon1993

Answer 10

well lets see if we flip that second fraction and change this to multiplication...we have

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

Answer 11

Lets factor that n^2 - 1

(n + 1)(n - 1)

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

Looks like we can cancel something ...

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

Answer 12

So now we have

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

Lets see what we can do to the top...well each term has a 'n' in it...lets factor that out...

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

Answer 13

Now we can cancel the 'n' on top by taking away 1 of the 'n's from the bottom

\[\large \frac{\cancel{n}(3n - 1)}{n\cancel{^2}(n - 1)}\]

So now we have

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

Anything else we can do?

Answer 14

Ok. I have a question. On the fraction top, how come you canceled out the n and the 2 on the bottom. Why not n? on the bottom?

Answer 15

Well because...we have

n^1 on top...
and
n^2 on the bottom...

If we want to take away 1 n from both of those...we have no 'n' left on top...and we have 1 n left on the bottom

Answer 16

Look at it like this

\[\large \frac{n}{n^2} = \frac{n}{n \times n} = \frac{\cancel{n}}{\cancel{n} \times n} = \frac{1}{n}\]

Answer 17

Aww. Ok. Got it!! Thanks for your BEST explanation!!!!

Answer 18

Anytime :)