Solve 2t/t-4-st/t^2-16

$$\frac { 2 t } { t - 4 } - \frac { s t } { t ^ { 2 } - 16 }$$

Evaluate

$$\frac{t\left(8+2t-s\right)}{t^{2}-16}$$

Short Solution Steps

Factor t^{2}-16.

$$\frac{2t}{t-4}-\frac{st}{\left(t-4\right)\left(t+4\right)}$$

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

$$\frac{2t\left(t+4\right)}{\left(t-4\right)\left(t+4\right)}-\frac{st}{\left(t-4\right)\left(t+4\right)}$$

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

$$\frac{2t\left(t+4\right)-st}{\left(t-4\right)\left(t+4\right)}$$

Do the multiplications in 2t\left(t+4\right)-st.

$$\frac{2t^{2}+8t-st}{\left(t-4\right)\left(t+4\right)}$$

Expand \left(t-4\right)\left(t+4\right).

$$\frac{2t^{2}+8t-st}{t^{2}-16}$$

Factor

$$\frac{t\left(8+2t-s\right)}{t^{2}-16}$$