Simplify (3k-27)/(k^2-8k-9)+(7k+14)/(3k^2+9k+6)

$\frac{3 k - 27}{k^{2} - 8 k - 9} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Simplify each term.

Factor $3$ out of $3 k - 27$ .

Factor $3$ out of $3 k$ .

$\frac{3 (k) - 27}{k^{2} - 8 k - 9} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Factor $3$ out of $- 27$ .

$\frac{3 k + 3 \cdot - 9}{k^{2} - 8 k - 9} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Factor $3$ out of $3 k + 3 \cdot - 9$ .

$\frac{3 (k - 9)}{k^{2} - 8 k - 9} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Factor $k^{2} - 8 k - 9$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 9$ and whose sum is $- 8$ .

$- 9, 1$

Write the factored form using these integers.

$\frac{3 (k - 9)}{(k - 9) (k + 1)} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Cancel the common factor of $k - 9$ .

Cancel the common factor.

$\frac{3 (k - 9)}{(k - 9) (k + 1)} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Rewrite the expression.

$\frac{3}{k + 1} + \frac{7 k + 14}{3 k^{2} + 9 k + 6}$

Factor $7$ out of $7 k + 14$ .

Factor $7$ out of $7 k$ .

$\frac{3}{k + 1} + \frac{7 (k) + 14}{3 k^{2} + 9 k + 6}$

Factor $7$ out of $14$ .

$\frac{3}{k + 1} + \frac{7 k + 7 \cdot 2}{3 k^{2} + 9 k + 6}$

Factor $7$ out of $7 k + 7 \cdot 2$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 k^{2} + 9 k + 6}$

Simplify the denominator.

Factor $3$ out of $3 k^{2} + 9 k + 6$ .

Factor $3$ out of $3 k^{2}$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k^{2}) + 9 k + 6}$

Factor $3$ out of $9 k$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k^{2}) + 3 (3 k) + 6}$

Factor $3$ out of $6$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 k^{2} + 3 (3 k) + 3 \cdot 2}$

Factor $3$ out of $3 k^{2} + 3 (3 k)$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k^{2} + 3 k) + 3 \cdot 2}$

Factor $3$ out of $3 (k^{2} + 3 k) + 3 \cdot 2$ .

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k^{2} + 3 k + 2)}$

Factor $k^{2} + 3 k + 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

Write the factored form using these integers.

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 ((k + 1) (k + 2))}$

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k + 1) (k + 2)}$

Cancel the common factor of $k + 2$ .

Cancel the common factor.

$\frac{3}{k + 1} + \frac{7 (k + 2)}{3 (k + 1) (k + 2)}$

Rewrite the expression.

$\frac{3}{k + 1} + \frac{7}{3 (k + 1)}$

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

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

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

Multiply $\frac{3}{k + 1}$ by $\frac{3}{3}$ .

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

Reorder the factors of $(k + 1) \cdot 3$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $3$ by $3$ .

$\frac{9 + 7}{3 (k + 1)}$

Add $9$ and $7$ .

$\frac{16}{3 (k + 1)}$

Simplify (3k-27)/(k^2-8k-9)+(7k+14)/(3k^2+9k+6)

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