Solve 3r/r+2+5t/r-2-40/r^2-4

$$\frac { 3 r } { r + 2 } + \frac { 5 t } { r - 2 } - \frac { 40 } { r ^ { 2 } - 4 }$$

Evaluate

$$\frac{3r^{2}+5rt-6r+10t-40}{r^{2}-4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of r+2 and r-2 is \left(r-2\right)\left(r+2\right). Multiply \frac{3r}{r+2} times \frac{r-2}{r-2}. Multiply \frac{5t}{r-2} times \frac{r+2}{r+2}.

$$\frac{3r\left(r-2\right)}{\left(r-2\right)\left(r+2\right)}+\frac{5t\left(r+2\right)}{\left(r-2\right)\left(r+2\right)}-\frac{40}{r^{2}-4}$$

Since \frac{3r\left(r-2\right)}{\left(r-2\right)\left(r+2\right)} and \frac{5t\left(r+2\right)}{\left(r-2\right)\left(r+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{3r\left(r-2\right)+5t\left(r+2\right)}{\left(r-2\right)\left(r+2\right)}-\frac{40}{r^{2}-4}$$

Do the multiplications in 3r\left(r-2\right)+5t\left(r+2\right).

$$\frac{3r^{2}-6r+5tr+10t}{\left(r-2\right)\left(r+2\right)}-\frac{40}{r^{2}-4}$$

Factor r^{2}-4.

$$\frac{3r^{2}-6r+5tr+10t}{\left(r-2\right)\left(r+2\right)}-\frac{40}{\left(r-2\right)\left(r+2\right)}$$

Since \frac{3r^{2}-6r+5tr+10t}{\left(r-2\right)\left(r+2\right)} and \frac{40}{\left(r-2\right)\left(r+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3r^{2}-6r+5tr+10t-40}{\left(r-2\right)\left(r+2\right)}$$

Expand \left(r-2\right)\left(r+2\right).

$$\frac{3r^{2}-6r+5tr+10t-40}{r^{2}-4}$$

Differentiate w.r.t. t

$$\frac{5}{r-2}$$