Question

Simplify the complex rational expression by writing it as division:

$\frac{b-{\displaystyle \frac{3b}{b+5}}}{{\displaystyle \frac{2}{b+5}}+{\displaystyle \frac{1}{b-5}}}$

Short answer

$\frac{\left(b^2+2b\right)\left(b-5\right)}{\left(3b-5\right)}$

Step-by-step solution

Step 1. Given information.

The following complex rational expression is given:

$\frac{b-{\displaystyle \frac{3b}{b+5}}}{{\displaystyle \frac{2}{b+5}+\frac{1}{b-5}}}$

Step 2. Simplification of the complex rational expression.

Simplify the numerator and the denominator by finding the common denominators:

$\frac{{\displaystyle \frac{b\left(b+5\right)}{\left(b+5\right)}}-{\displaystyle \frac{3b}{b+5}}}{{\displaystyle \frac{2\left(b-5\right)}{\left(b+5\right)\left(b-5\right)}+\frac{1\left(b+5\right)}{\left(b-5\right)\left(b+5\right)}}}$

Simplify the numerator and the denominator:

$$\begin{gathered} \frac{{\displaystyle \frac{b\left(b+5\right)-3b}{\left(b+5\right)}}}{{\displaystyle \frac{2\left(b-5\right)+\left(b+5\right)}{\left(b+5\right)\left(b-5\right)}}} \\ \frac{{\displaystyle \frac{b^2+2b}{\left(b+5\right)}}}{{\displaystyle \frac{3b-5}{\left(b+5\right)\left(b-5\right)}}} \end{gathered}$$

Write as a division problem:

$\frac{b^2+2b}{\left(b+5\right)}÷\frac{3b-5}{\left(b+5\right)\left(b-5\right)}$

Write as the product of the numerator and the reciprocal of the denominator:

$\frac{b^2+2b}{\left(b+5\right)}.\frac{\left(b+5\right)\left(b-5\right)}{3b-5}$

Cancel the common factors from the numerator and the denominator and simplify:

$\frac{\left(b^2+2b\right)\left(b-5\right)}{\left(3b-5\right)}$