Question

In the following exercises, simplify.

$\frac{{\displaystyle \frac{5}{b^2-6b-27}}}{{\displaystyle \frac{3}{b+3}}+{\displaystyle \frac{1}{b-9}}}$

Short answer

The simplified expression is$\frac{5}{4b}$

Step-by-step solution

Step 1. Given information

The given expression is

$\frac{{\displaystyle \frac{5}{b^2-6b-27}}}{{\displaystyle \frac{3}{b+3}+\frac{1}{b-9}}}$

Step 2. Simplify the expression

$$\begin{gathered} \frac{{\displaystyle \frac{5}{b^2-6b-27}}}{{\displaystyle \frac{3}{b+3}+\frac{1}{b-9}}}=\frac{\left(\cancel{b+3}\right)\left(\cancel{b-9}\right)\left({\displaystyle \frac{5}{\left(\cancel{b+3}\right)\left(\cancel{b-9}\right)}}\right)}{\left(\cancel{b+3}\right)\left(\cancel{b-9}\right)\left({\displaystyle \frac{3}{\left(\cancel{b-9}\right)}}+{\displaystyle \frac{1}{\left(\cancel{b+3}\right)}}\right)} \\ =\frac{5}{3\left(b+3\right)+\left(b-9\right)} \\ =\frac{5}{4b} \end{gathered}$$

The simplified expression is$\frac{5}{4b}$

Step 3. The solution 

The simplified expression is$\frac{5}{4b}$