Solve (2k-1/k+1-2k+1/k-1):(1+1/k-1)

$$( \frac { 2 k - 1 } { k + 1 } - \frac { 2 k + 1 } { k - 1 } ) : ( 1 + \frac { 1 } { k - 1 } )$$

Evaluate

$$-\frac{6}{k+1}$$

Short Solution Steps

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

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

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

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

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

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

Combine like terms in 2k^{2}-2k-k+1-2k^{2}-2k-k-1.

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

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

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

Since \frac{k-1}{k-1} and \frac{1}{k-1} have the same denominator, add them by adding their numerators.

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

Combine like terms in k-1+1.

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

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

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

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

$$\frac{-6}{k+1}$$

Expand

$$-\frac{6}{k+1}$$

Short Solution Steps

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

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

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

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

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

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

Combine like terms in 2k^{2}-2k-k+1-2k^{2}-2k-k-1.

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

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

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

Since \frac{k-1}{k-1} and \frac{1}{k-1} have the same denominator, add them by adding their numerators.

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

Combine like terms in k-1+1.

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

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

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

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

$$\frac{-6}{k+1}$$