Solve 3k/3k-6-2k/k+3

$$\frac { 3 k } { 3 k - 6 } - \frac { 2 k } { k + 3 }$$

Evaluate

$$\frac{k\left(7-k\right)}{\left(k-2\right)\left(k+3\right)}$$

Short Solution Steps

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

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

Cancel out 3 in both numerator and denominator.

$$\frac{k}{k-2}-\frac{2k}{k+3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of k-2 and k+3 is \left(k-2\right)\left(k+3\right). Multiply \frac{k}{k-2} times \frac{k+3}{k+3}. Multiply \frac{2k}{k+3} times \frac{k-2}{k-2}.

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

Since \frac{k\left(k+3\right)}{\left(k-2\right)\left(k+3\right)} and \frac{2k\left(k-2\right)}{\left(k-2\right)\left(k+3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in k\left(k+3\right)-2k\left(k-2\right).

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

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

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

Expand \left(k-2\right)\left(k+3\right).

$$\frac{-k^{2}+7k}{k^{2}+k-6}$$

Factor

$$\frac{k\left(7-k\right)}{\left(k-2\right)\left(k+3\right)}$$