Solve k^2-6k-27/k+3+8k^2-72k/6k-6

$$\frac { k ^ { 2 } - 6 k - 27 } { k + 3 } + \frac { 8 k ^ { 2 } - 72 k } { 6 k - 6 }$$

Evaluate

$$\frac{\left(k-9\right)\left(7k-3\right)}{3\left(k-1\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{k^{2}-6k-27}{k+3}.

$$\frac{\left(k-9\right)\left(k+3\right)}{k+3}+\frac{8k^{2}-72k}{6k-6}$$

Cancel out k+3 in both numerator and denominator.

$$k-9+\frac{8k^{2}-72k}{6k-6}$$

Factor the expressions that are not already factored in \frac{8k^{2}-72k}{6k-6}.

$$k-9+\frac{8k\left(k-9\right)}{6\left(k-1\right)}$$

Cancel out 2 in both numerator and denominator.

$$k-9+\frac{4k\left(k-9\right)}{3\left(k-1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply k-9 times \frac{3\left(k-1\right)}{3\left(k-1\right)}.

$$\frac{\left(k-9\right)\times 3\left(k-1\right)}{3\left(k-1\right)}+\frac{4k\left(k-9\right)}{3\left(k-1\right)}$$

Since \frac{\left(k-9\right)\times 3\left(k-1\right)}{3\left(k-1\right)} and \frac{4k\left(k-9\right)}{3\left(k-1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k-9\right)\times 3\left(k-1\right)+4k\left(k-9\right)}{3\left(k-1\right)}$$

Do the multiplications in \left(k-9\right)\times 3\left(k-1\right)+4k\left(k-9\right).

$$\frac{3k^{2}-3k-27k+27+4k^{2}-36k}{3\left(k-1\right)}$$

Combine like terms in 3k^{2}-3k-27k+27+4k^{2}-36k.

$$\frac{7k^{2}-66k+27}{3\left(k-1\right)}$$

Expand 3\left(k-1\right).

$$\frac{7k^{2}-66k+27}{3k-3}$$

Expand

$$\frac{7k^{2}-66k+27}{3\left(k-1\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{k^{2}-6k-27}{k+3}.

$$\frac{\left(k-9\right)\left(k+3\right)}{k+3}+\frac{8k^{2}-72k}{6k-6}$$

Cancel out k+3 in both numerator and denominator.

$$k-9+\frac{8k^{2}-72k}{6k-6}$$

Factor the expressions that are not already factored in \frac{8k^{2}-72k}{6k-6}.

$$k-9+\frac{8k\left(k-9\right)}{6\left(k-1\right)}$$

Cancel out 2 in both numerator and denominator.

$$k-9+\frac{4k\left(k-9\right)}{3\left(k-1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply k-9 times \frac{3\left(k-1\right)}{3\left(k-1\right)}.

$$\frac{\left(k-9\right)\times 3\left(k-1\right)}{3\left(k-1\right)}+\frac{4k\left(k-9\right)}{3\left(k-1\right)}$$

Since \frac{\left(k-9\right)\times 3\left(k-1\right)}{3\left(k-1\right)} and \frac{4k\left(k-9\right)}{3\left(k-1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k-9\right)\times 3\left(k-1\right)+4k\left(k-9\right)}{3\left(k-1\right)}$$

Do the multiplications in \left(k-9\right)\times 3\left(k-1\right)+4k\left(k-9\right).

$$\frac{3k^{2}-3k-27k+27+4k^{2}-36k}{3\left(k-1\right)}$$

Combine like terms in 3k^{2}-3k-27k+27+4k^{2}-36k.

$$\frac{7k^{2}-66k+27}{3\left(k-1\right)}$$

Expand 3\left(k-1\right).

$$\frac{7k^{2}-66k+27}{3k-3}$$