What is (n^2-9)/(n+3)*n/(2n-6) ?

The solution to (n^2-9)/(n+3)*n/(2n-6) is n/2

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simplify (n^2-9)/(n+3)*n/(2n-6)

simplify

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

\frac{n}{2}

Solution steps

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

Cancel

\frac{n ^{2} - 9}{n + 3} : n - 3

= (n - 3) \frac{n}{2 n - 6}

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( n - 3 ) n}{2 n - 6}

Factor

2 n - 6 : 2 (n - 3)

= \frac{( n - 3 ) n}{2 ( n - 3 )}

Cancel the common factor:

n - 3

= \frac{n}{2}

15 x - 20 = 6 x - (x + 2) + (- x + 3)

x^{2} - \frac{3}{2} x - \frac{27}{16} = 0

\frac{5}{3} (x + \frac{1}{2}) = \frac{95}{6}

\frac{π}{2} + \frac{2 π}{3}

\frac{7 π}{3} - \frac{π}{3}