Solve for k (k-15)/(k+1)=(-4k-12)/(k^2-9)

$\frac{k - 15}{k + 1} = \frac{- 4 k - 12}{k^{2} - 9}$

Step 1

Simplify both sides.

Factor $4$ out of $- 4 k - 12$ .

Factor $4$ out of $- 4 k$ .

$\frac{k - 15}{k + 1} = \frac{4 (- k) - 12}{k^{2} - 9}$

Factor $4$ out of $- 12$ .

$\frac{k - 15}{k + 1} = \frac{4 (- k) + 4 (- 3)}{k^{2} - 9}$

Factor $4$ out of $4 (- k) + 4 (- 3)$ .

$\frac{k - 15}{k + 1} = \frac{4 (- k - 3)}{k^{2} - 9}$

Simplify the denominator.

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

$\frac{k - 15}{k + 1} = \frac{4 (- k - 3)}{k^{2} - 3^{2}}$

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

$\frac{k - 15}{k + 1} = \frac{4 (- k - 3)}{(k + 3) (k - 3)}$

Cancel the common factor of $- k - 3$ and $k + 3$ .

Factor $- 1$ out of $- k$ .

$\frac{k - 15}{k + 1} = \frac{4 (- (k) - 3)}{(k + 3) (k - 3)}$

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

$\frac{k - 15}{k + 1} = \frac{4 (- (k) - 1 (3))}{(k + 3) (k - 3)}$

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

$\frac{k - 15}{k + 1} = \frac{4 (- (k + 3))}{(k + 3) (k - 3)}$

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

$\frac{k - 15}{k + 1} = \frac{4 (- 1 (k + 3))}{(k + 3) (k - 3)}$

Cancel the common factor.

$\frac{k - 15}{k + 1} = \frac{4 (- 1 (k + 3))}{(k + 3) (k - 3)}$

Rewrite the expression.

$\frac{k - 15}{k + 1} = \frac{4 \cdot (- 1)}{k - 3}$

Multiply $4$ by $- 1$ .

$\frac{k - 15}{k + 1} = \frac{- 4}{k - 3}$

Move the negative in front of the fraction.

$\frac{k - 15}{k + 1} = - \frac{4}{k - 3}$

Step 2

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(k - 15) (k - 3) = (k + 1) \cdot - 4$

Step 3

Solve the equation for $k$ .

Simplify $(k - 15) (k - 3)$ .

Start simplifying.

$0 + 0 + (k - 15) (k - 3) = (k + 1) \cdot - 4$

Simplify by adding zeros.

$(k - 15) (k - 3) = (k + 1) \cdot - 4$

Expand $(k - 15) (k - 3)$ using the FOIL Method.

Apply the distributive property.

$k (k - 3) - 15 (k - 3) = (k + 1) \cdot - 4$

Apply the distributive property.

$k \cdot k + k \cdot - 3 - 15 (k - 3) = (k + 1) \cdot - 4$

Apply the distributive property.

$k \cdot k + k \cdot - 3 - 15 k - 15 \cdot - 3 = (k + 1) \cdot - 4$

Simplify and combine like terms.

Simplify each term.

Multiply $k$ by $k$ .

$k^{2} + k \cdot - 3 - 15 k - 15 \cdot - 3 = (k + 1) \cdot - 4$

Move $- 3$ to the left of $k$ .

$k^{2} - 3 \cdot k - 15 k - 15 \cdot - 3 = (k + 1) \cdot - 4$

Multiply $- 15$ by $- 3$ .

$k^{2} - 3 k - 15 k + 45 = (k + 1) \cdot - 4$

Subtract $15 k$ from $- 3 k$ .

$k^{2} - 18 k + 45 = (k + 1) \cdot - 4$

Simplify $(k + 1) \cdot - 4$ .

Apply the distributive property.

$k^{2} - 18 k + 45 = k \cdot - 4 + 1 \cdot - 4$

Simplify the expression.

Move $- 4$ to the left of $k$ .

$k^{2} - 18 k + 45 = - 4 \cdot k + 1 \cdot - 4$

Multiply $- 4$ by $1$ .

$k^{2} - 18 k + 45 = - 4 k - 4$

Move all terms containing $k$ to the left side of the equation.

Add $4 k$ to both sides of the equation.

$k^{2} - 18 k + 45 + 4 k = - 4$

Add $- 18 k$ and $4 k$ .

$k^{2} - 14 k + 45 = - 4$

Add $4$ to both sides of the equation.

$k^{2} - 14 k + 45 + 4 = 0$

Add $45$ and $4$ .

$k^{2} - 14 k + 49 = 0$

Factor using the perfect square rule.

Rewrite $49$ as $7^{2}$ .

$k^{2} - 14 k + 7^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$14 k = 2 \cdot k \cdot 7$

Rewrite the polynomial.

$k^{2} - 2 \cdot k \cdot 7 + 7^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = k$ and $b = 7$ .

${(k - 7)}^{2} = 0$

Set the $k - 7$ equal to $0$ .

$k - 7 = 0$

Add $7$ to both sides of the equation.

$k = 7$

Solve for k (k-15)/(k+1)=(-4k-12)/(k^2-9)

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