Respuesta :

LammettHash

First some rewriting:

[tex]\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)}}[/tex]

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

[tex]\dfrac{n+3}{2(n-3)}\div\dfrac{n+3}{3(n-3)}=\dfrac{\frac12}{\frac13}[/tex]

Then

[tex]\dfrac{\frac12}{\frac13}=\dfrac12\cdot\dfrac31=\dfrac32[/tex]

Answer:

[tex]\frac{3}{2}[/tex]

Step-by-step explanation:

[tex]\frac{n+3}{2n-6} divide \frac{n+3}{2n-9}[/tex]

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

[tex]\frac{n+3}{2n-6} divide \frac{n+3}{2n-9}[/tex]

[tex]\frac{n+3}{2n-6} \cdot \frac{3n-9}{n+3}[/tex]

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

[tex]2n-6=2(n-3)[/tex]

[tex]3n-9=3(n-3)[/tex]

Replace the factors

[tex]\frac{n+3}{2(n-3)} \cdot \frac{3(n-3)}{n+3}[/tex]

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

[tex]\frac{3}{2}[/tex]

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