Solve 2k/2k+7-6/5k-1

$$\frac { 2 k } { 2 k + 7 } - \frac { 6 } { 5 k - 1 }$$

Evaluate

$$\frac{2\left(5k^{2}-7k-21\right)}{\left(5k-1\right)\left(2k+7\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2k+7 and 5k-1 is \left(5k-1\right)\left(2k+7\right). Multiply \frac{2k}{2k+7} times \frac{5k-1}{5k-1}. Multiply \frac{6}{5k-1} times \frac{2k+7}{2k+7}.

$$\frac{2k\left(5k-1\right)}{\left(5k-1\right)\left(2k+7\right)}-\frac{6\left(2k+7\right)}{\left(5k-1\right)\left(2k+7\right)}$$

Since \frac{2k\left(5k-1\right)}{\left(5k-1\right)\left(2k+7\right)} and \frac{6\left(2k+7\right)}{\left(5k-1\right)\left(2k+7\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2k\left(5k-1\right)-6\left(2k+7\right)}{\left(5k-1\right)\left(2k+7\right)}$$

Do the multiplications in 2k\left(5k-1\right)-6\left(2k+7\right).

$$\frac{10k^{2}-2k-12k-42}{\left(5k-1\right)\left(2k+7\right)}$$

Combine like terms in 10k^{2}-2k-12k-42.

$$\frac{10k^{2}-14k-42}{\left(5k-1\right)\left(2k+7\right)}$$

Expand \left(5k-1\right)\left(2k+7\right).

$$\frac{10k^{2}-14k-42}{10k^{2}+33k-7}$$

Factor

$$\frac{10\left(k-\frac{7-\sqrt{469}}{10}\right)\left(k-\frac{\sqrt{469}+7}{10}\right)}{\left(5k-1\right)\left(2k+7\right)}$$