Question

What is the product?

StartFraction 4 n Over 4 n minus 4 EndFraction times StartFraction n minus 1 Over n + 1 EndFraction

Answer 1

`(4n)/(4n-4) (n-1)/(n+1)`

We can take common factor of 4 for the term
`(4n)/(4n-4)` and we got:

`(4)/(4) (n)/(n-1) (n-1)/(n+1)`

Now we can cancel the n-1 in the denominator and the n-1 in the numerator and we got:

`1 (n)/(n+1)`

And the final answer would be:

`(n)/(n+1)`

Explanation:

For this case we have the following product given:

`(4n)/(4n-4) (n-1)/(n+1)`

We can take common factor of 4 for the term
`(4n)/(4n-4)` and we got:

`(4)/(4) (n)/(n-1) (n-1)/(n+1)`

Now we can cancel the n-1 in the denominator and the n-1 in the numerator and we got:

`1 (n)/(n+1)`

And the final answer would be:

`(n)/(n+1)`

Answer 2

`(n)/(n+1)`

Explanation:

Given the expression
`(4n)/(4n-4) *(n-1)/(n+1)`

First we will factor out 4 from 4n-4 at the denominator to have;

`= (4n)/(4(n-1)) * (n-1)/(n+1)`

Cancelling out the n-1 from both numerator and denominator we have;

`= (4n)/(4(n+1)) \\= (n)/(n+1)`

n/n+1 gives the required product of the expression