Solve k+5/k+1+8/k^2-1=

$$\frac { k + 5 } { k + 1 } + \frac { 8 } { k ^ { 2 } - 1 } =$$

Evaluate

$$\frac{k+3}{k-1}$$

Short Solution Steps

Factor k^{2}-1.

$$\frac{k+5}{k+1}+\frac{8}{\left(k-1\right)\left(k+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k+1 and \left(k-1\right)\left(k+1\right) is \left(k-1\right)\left(k+1\right). Multiply \frac{k+5}{k+1} times \frac{k-1}{k-1}.

$$\frac{\left(k+5\right)\left(k-1\right)}{\left(k-1\right)\left(k+1\right)}+\frac{8}{\left(k-1\right)\left(k+1\right)}$$

Since \frac{\left(k+5\right)\left(k-1\right)}{\left(k-1\right)\left(k+1\right)} and \frac{8}{\left(k-1\right)\left(k+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k+5\right)\left(k-1\right)+8}{\left(k-1\right)\left(k+1\right)}$$

Do the multiplications in \left(k+5\right)\left(k-1\right)+8.

$$\frac{k^{2}-k+5k-5+8}{\left(k-1\right)\left(k+1\right)}$$

Combine like terms in k^{2}-k+5k-5+8.

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

Factor the expressions that are not already factored in \frac{k^{2}+4k+3}{\left(k-1\right)\left(k+1\right)}.

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

Cancel out k+1 in both numerator and denominator.

$$\frac{k+3}{k-1}$$

Expand

$$\frac{k+3}{k-1}$$

Short Solution Steps

Factor k^{2}-1.

$$\frac{k+5}{k+1}+\frac{8}{\left(k-1\right)\left(k+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k+1 and \left(k-1\right)\left(k+1\right) is \left(k-1\right)\left(k+1\right). Multiply \frac{k+5}{k+1} times \frac{k-1}{k-1}.

$$\frac{\left(k+5\right)\left(k-1\right)}{\left(k-1\right)\left(k+1\right)}+\frac{8}{\left(k-1\right)\left(k+1\right)}$$

Since \frac{\left(k+5\right)\left(k-1\right)}{\left(k-1\right)\left(k+1\right)} and \frac{8}{\left(k-1\right)\left(k+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k+5\right)\left(k-1\right)+8}{\left(k-1\right)\left(k+1\right)}$$

Do the multiplications in \left(k+5\right)\left(k-1\right)+8.

$$\frac{k^{2}-k+5k-5+8}{\left(k-1\right)\left(k+1\right)}$$

Combine like terms in k^{2}-k+5k-5+8.

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

Factor the expressions that are not already factored in \frac{k^{2}+4k+3}{\left(k-1\right)\left(k+1\right)}.

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

Cancel out k+1 in both numerator and denominator.

$$\frac{k+3}{k-1}$$