Simplify 3/(n^2-9)+7/(3-n)

$\frac{3}{n^{2} - 9} + \frac{7}{3 - n}$

Step 1

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$\frac{3}{n^{2} - 3^{2}} + \frac{7}{3 - n}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = n$ and $b = 3$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{7}{3 - n}$

Step 2

Simplify with factoring out.

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{7}{- 1 (- 3) - n}$

Factor $- 1$ out of $- n$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{7}{- 1 (- 3) - (n)}$

Factor $- 1$ out of $- 1 (- 3) - (n)$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{7}{- 1 (- 3 + n)}$

Simplify the expression.

Move a negative from the denominator of $\frac{7}{- 1 (- 3 + n)}$ to the numerator.

$\frac{3}{(n + 3) (n - 3)} + \frac{- 1 \cdot 7}{- 3 + n}$

Reorder terms.

$\frac{3}{(n + 3) (n - 3)} + \frac{- 1 \cdot 7}{n - 3}$

Step 3

To write $\frac{- 1 \cdot 7}{n - 3}$ as a fraction with a common denominator, multiply by $\frac{n + 3}{n + 3}$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{- 1 \cdot 7}{n - 3} \cdot \frac{n + 3}{n + 3}$

Step 4

Write each expression with a common denominator of $(n + 3) (n - 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{- 1 \cdot 7}{n - 3}$ by $\frac{n + 3}{n + 3}$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{- 1 \cdot 7 (n + 3)}{(n - 3) (n + 3)}$

Reorder the factors of $(n - 3) (n + 3)$ .

$\frac{3}{(n + 3) (n - 3)} + \frac{- 1 \cdot 7 (n + 3)}{(n + 3) (n - 3)}$

Step 5

Combine the numerators over the common denominator.

$\frac{3 - 1 \cdot 7 (n + 3)}{(n + 3) (n - 3)}$

Step 6

Simplify the numerator.

Multiply $- 1$ by $7$ .

$\frac{3 - 7 (n + 3)}{(n + 3) (n - 3)}$

Apply the distributive property.

$\frac{3 - 7 n - 7 \cdot 3}{(n + 3) (n - 3)}$

Multiply $- 7$ by $3$ .

$\frac{3 - 7 n - 21}{(n + 3) (n - 3)}$

Subtract $21$ from $3$ .

$\frac{- 7 n - 18}{(n + 3) (n - 3)}$

Step 7

Simplify with factoring out.

Factor $- 1$ out of $- 7 n$ .

$\frac{- (7 n) - 18}{(n + 3) (n - 3)}$

Rewrite $- 18$ as $- 1 (18)$ .

$\frac{- (7 n) - 1 (18)}{(n + 3) (n - 3)}$

Factor $- 1$ out of $- (7 n) - 1 (18)$ .

$\frac{- (7 n + 18)}{(n + 3) (n - 3)}$

Simplify the expression.

Rewrite $- (7 n + 18)$ as $- 1 (7 n + 18)$ .

$\frac{- 1 (7 n + 18)}{(n + 3) (n - 3)}$

Move the negative in front of the fraction.

$- \frac{7 n + 18}{(n + 3) (n - 3)}$

Simplify 3/(n^2-9)+7/(3-n)

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