What is n/(n+3)+(12n)/(n-3)-9/(n^2-9) ?

The solution to n/(n+3)+(12n)/(n-3)-9/(n^2-9) is (13n^2+33n-9)/((n+3)(n-3))

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simplify n/(n+3)+(12n)/(n-3)-9/(n^2-9)

simplify

\frac{n}{n + 3} + \frac{12 n}{n - 3} - \frac{9}{n ^{2} - 9}

\frac{13 n ^{2} + 33 n - 9}{( n + 3 ) ( n - 3 )}

Solution steps

\frac{n}{n + 3} + \frac{12 n}{n - 3} - \frac{9}{n ^{2} - 9}

Least Common Multiplier of

n + 3, n - 3, n^{2} - 9 : (n + 3) (n - 3)

Adjust Fractions based on the LCM

= \frac{n ( n - 3 )}{( n + 3 ) ( n - 3 )} + \frac{12 n ( n + 3 )}{( n + 3 ) ( n - 3 )} - \frac{9}{( n + 3 ) ( n - 3 )}

Apply the fraction rule:

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

= \frac{n ( n - 3 ) + 12 n ( n + 3 ) - 9}{( n + 3 ) ( n - 3 )}

Simplify

n (n - 3) + 12 n (n + 3) - 9 : 13 n^{2} + 33 n - 9

= \frac{13 n ^{2} + 33 n - 9}{( n + 3 ) ( n - 3 )}

3 b^{2} - 6 b

4 (x^{2} + 14 x + 49) - 1 = 0

\frac{5}{19} \cdot 6

(x x)^{2}

2 x + 3 \geq 13

What is n/(n+3)+(12n)/(n-3)-9/(n^2-9) ?
The solution to n/(n+3)+(12n)/(n-3)-9/(n^2-9) is (13n^2+33n-9)/((n+3)(n-3))