Solve k/3k-8-4/k+2

$$\frac { k } { 3 k - 8 } - \frac { 4 } { k + 2 }$$

Evaluate

$$\frac{k^{2}-10k+32}{\left(3k-8\right)\left(k+2\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3k-8 and k+2 is \left(3k-8\right)\left(k+2\right). Multiply \frac{k}{3k-8} times \frac{k+2}{k+2}. Multiply \frac{4}{k+2} times \frac{3k-8}{3k-8}.

$$\frac{k\left(k+2\right)}{\left(3k-8\right)\left(k+2\right)}-\frac{4\left(3k-8\right)}{\left(3k-8\right)\left(k+2\right)}$$

Since \frac{k\left(k+2\right)}{\left(3k-8\right)\left(k+2\right)} and \frac{4\left(3k-8\right)}{\left(3k-8\right)\left(k+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{k\left(k+2\right)-4\left(3k-8\right)}{\left(3k-8\right)\left(k+2\right)}$$

Do the multiplications in k\left(k+2\right)-4\left(3k-8\right).

$$\frac{k^{2}+2k-12k+32}{\left(3k-8\right)\left(k+2\right)}$$

Combine like terms in k^{2}+2k-12k+32.

$$\frac{k^{2}-10k+32}{\left(3k-8\right)\left(k+2\right)}$$

Expand \left(3k-8\right)\left(k+2\right).

$$\frac{k^{2}-10k+32}{3k^{2}-2k-16}$$

Differentiate w.r.t. k

$$\frac{28\left(k^{2}-8k+8\right)}{9k^{4}-12k^{3}-92k^{2}+64k+256}$$