Solve 5/n-1-1/n+1/frac{1{n+1}-2/n-1}

$$\frac { \frac { 5 } { n - 1 } - \frac { 1 } { n + 1 } } { \frac { 1 } { n + 1 } - \frac { 2 } { n - 1 } }$$

Evaluate

$$-\frac{2\left(2n+3\right)}{n+3}$$

Short Solution Steps

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

$$\frac{\frac{5\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}-\frac{n-1}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

Since \frac{5\left(n+1\right)}{\left(n-1\right)\left(n+1\right)} and \frac{n-1}{\left(n-1\right)\left(n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{5\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

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

$$\frac{\frac{5n+5-n+1}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

Combine like terms in 5n+5-n+1.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n+1 and n-1 is \left(n-1\right)\left(n+1\right). Multiply \frac{1}{n+1} times \frac{n-1}{n-1}. Multiply \frac{2}{n-1} times \frac{n+1}{n+1}.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1}{\left(n-1\right)\left(n+1\right)}-\frac{2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}}$$

Since \frac{n-1}{\left(n-1\right)\left(n+1\right)} and \frac{2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1-2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}}$$

Do the multiplications in n-1-2\left(n+1\right).

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1-2n-2}{\left(n-1\right)\left(n+1\right)}}$$

Combine like terms in n-1-2n-2.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{-n-3}{\left(n-1\right)\left(n+1\right)}}$$

Divide \frac{4n+6}{\left(n-1\right)\left(n+1\right)} by \frac{-n-3}{\left(n-1\right)\left(n+1\right)} by multiplying \frac{4n+6}{\left(n-1\right)\left(n+1\right)} by the reciprocal of \frac{-n-3}{\left(n-1\right)\left(n+1\right)}.

$$\frac{\left(4n+6\right)\left(n-1\right)\left(n+1\right)}{\left(n-1\right)\left(n+1\right)\left(-n-3\right)}$$

Cancel out \left(n-1\right)\left(n+1\right) in both numerator and denominator.

$$\frac{4n+6}{-n-3}$$

Expand

$$-\frac{2\left(2n+3\right)}{n+3}$$

Short Solution Steps

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

$$\frac{\frac{5\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}-\frac{n-1}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

Since \frac{5\left(n+1\right)}{\left(n-1\right)\left(n+1\right)} and \frac{n-1}{\left(n-1\right)\left(n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{5\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

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

$$\frac{\frac{5n+5-n+1}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

Combine like terms in 5n+5-n+1.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{1}{n+1}-\frac{2}{n-1}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n+1 and n-1 is \left(n-1\right)\left(n+1\right). Multiply \frac{1}{n+1} times \frac{n-1}{n-1}. Multiply \frac{2}{n-1} times \frac{n+1}{n+1}.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1}{\left(n-1\right)\left(n+1\right)}-\frac{2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}}$$

Since \frac{n-1}{\left(n-1\right)\left(n+1\right)} and \frac{2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1-2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)}}$$

Do the multiplications in n-1-2\left(n+1\right).

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{n-1-2n-2}{\left(n-1\right)\left(n+1\right)}}$$

Combine like terms in n-1-2n-2.

$$\frac{\frac{4n+6}{\left(n-1\right)\left(n+1\right)}}{\frac{-n-3}{\left(n-1\right)\left(n+1\right)}}$$

Divide \frac{4n+6}{\left(n-1\right)\left(n+1\right)} by \frac{-n-3}{\left(n-1\right)\left(n+1\right)} by multiplying \frac{4n+6}{\left(n-1\right)\left(n+1\right)} by the reciprocal of \frac{-n-3}{\left(n-1\right)\left(n+1\right)}.

$$\frac{\left(4n+6\right)\left(n-1\right)\left(n+1\right)}{\left(n-1\right)\left(n+1\right)\left(-n-3\right)}$$

Cancel out \left(n-1\right)\left(n+1\right) in both numerator and denominator.

$$\frac{4n+6}{-n-3}$$