Question

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

Answer 1

First some rewriting:

$\dfrac{n+3}{2n-6}\div\dfrac{n+3}{3n-9}=\dfrac{n+3}{2(n-3)}\div\dfrac{n+3}{3(n-3)}=\dfrac{\frac{n+3}{2(n-3)}}{\frac{n+3}{3(n-3)}}$

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

$\dfrac{n+3}{2(n-3)}\div\dfrac{n+3}{3(n-3)}=\dfrac{\frac12}{\frac13}$

Then

$\dfrac{\frac12}{\frac13}=\dfrac12\cdot\dfrac31=\dfrac32$

Answer 2

Answer:

$\frac{3}{2}$

Step-by-step explanation:

$\frac{n+3}{2n-6} divide \frac{n+3}{2n-9}$

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

$\frac{n+3}{2n-6} divide \frac{n+3}{2n-9}$

$\frac{n+3}{2n-6} \cdot \frac{3n-9}{n+3}$

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

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

$3n-9=3(n-3)$

Replace the factors

$\frac{n+3}{2(n-3)} \cdot \frac{3(n-3)}{n+3}$

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

$\frac{3}{2}$