Solve d^2-3d-10/d^2-d-20+d^2-4/d^2+4d=

$$\frac { d ^ { 2 } - 3 d - 10 } { d ^ { 2 } - d - 20 } + \frac { d ^ { 2 } - 4 } { d ^ { 2 } + 4 d } =$$

Evaluate

$$\frac{2\left(d-1\right)\left(d+2\right)}{d\left(d+4\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{d^{2}-3d-10}{d^{2}-d-20}.

$$\frac{\left(d-5\right)\left(d+2\right)}{\left(d-5\right)\left(d+4\right)}+\frac{d^{2}-4}{d^{2}+4d}$$

Cancel out d-5 in both numerator and denominator.

$$\frac{d+2}{d+4}+\frac{d^{2}-4}{d^{2}+4d}$$

Factor d^{2}+4d.

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

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

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

Since \frac{\left(d+2\right)d}{d\left(d+4\right)} and \frac{d^{2}-4}{d\left(d+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(d+2\right)d+d^{2}-4.

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

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

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

Expand d\left(d+4\right).

$$\frac{2d^{2}+2d-4}{d^{2}+4d}$$

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{d^{2}-3d-10}{d^{2}-d-20}.

$$\frac{\left(d-5\right)\left(d+2\right)}{\left(d-5\right)\left(d+4\right)}+\frac{d^{2}-4}{d^{2}+4d}$$

Cancel out d-5 in both numerator and denominator.

$$\frac{d+2}{d+4}+\frac{d^{2}-4}{d^{2}+4d}$$

Factor d^{2}+4d.

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

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

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

Since \frac{\left(d+2\right)d}{d\left(d+4\right)} and \frac{d^{2}-4}{d\left(d+4\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(d+2\right)d+d^{2}-4.

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

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

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

Expand d\left(d+4\right).

$$\frac{2d^{2}+2d-4}{d^{2}+4d}$$