Question

In the following exercise, perform the indicated operation:

$\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{d^2-d-20}$

Short answer

$\frac{6d\left(d+3\right)}{\left(d-5\right)\left(d+4\right)}$

Step-by-step solution

Step 1. Given information.

The following expression is given:

$\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{d^2-d-20}$

Step 2. Performing the indicated operation.

Factorise the denominators (where applicable):

$\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{\left(d-5\right)\left(d+4\right)}$

Find out the LCD:

$\left(d-5\right)\left(d+4\right)$

Rewrite the expression with denominators equal to the LCD:

$\frac{6d\left(d+4\right)}{\left(d-5\right)\left(d+4\right)}+\frac{1\left(d-5\right)}{\left(d+4\right)\left(d-5\right)}-\frac{7d-5}{\left(d-5\right)\left(d+4\right)}$

Simplify:

$$\begin{gathered} \frac{6d^2+24d+d-5-7d+5}{\left(d-5\right)\left(d+4\right)} \\ \frac{6d^2+18d}{\left(d-5\right)\left(d+4\right)} \end{gathered}$$

Simplify:

$\frac{6d\left(d+3\right)}{\left(d-5\right)\left(d+4\right)}$