Question

What is the quotient? n+3/2n-6 divided by n+3/3n-9

Answer 1

First some rewriting:

`(n+3)/(2n-6)/(n+3)/(3n-9)=(n+3)/(2(n-3))/(n+3)/(3(n-3))=((n+3)/(2(n-3)))/((n+3)/(3(n-3)))`

We see that the numerators of both fractions
`(n+3)` and a term in the denominators
`(n-3)` will cancel:

`(n+3)/(2(n-3))/(n+3)/(3(n-3))=(\frac12)/(\frac13)`

Then

`(\frac12)/(\frac13)=\frac12\cdot\frac31=\frac32`

Answer 2

`(3)/(2)`

Explanation:

`(n+3)/(2n-6) divide (n+3)/(2n-9)`

To remove the division symbol, take reciprocal of second fraction and put multiplication symbol inbetween

`(n+3)/(2n-6) divide (n+3)/(2n-9)`

`(n+3)/(2n-6) \cdot (3n-9)/(n+3)`

Now we factor 2n-6 and 3n-9

`2n-6=2(n-3)`

`3n-9=3(n-3)`

Replace the factors

`(n+3)/(2(n-3)) \cdot (3(n-3))/(n+3)`

cancel out n+3 and n-3 at the numerator and denominator

`(3)/(2)`