Question

Simplify

In the following exercises, use either method.

$\frac{{\displaystyle \frac{y}{y+3}}}{2+{\displaystyle \frac{1}{y-3}}}$.

Short answer

The simplified expression for$\frac{{\displaystyle \frac{y}{y+3}}}{2+{\displaystyle \frac{1}{y-3}}}$is,$\frac{y^2-3y}{\left(y+3\right)\left(2y-5\right)}$.

Step-by-step solution

Step 1. Given information 

$\frac{{\displaystyle \frac{y}{y+3}}}{2+{\displaystyle \frac{1}{y-3}}}$.

We need to simplify the given expression.

Step 2.Find the LCD of all fractions in the complex rational expression.

The LCD is, $\left(y+3\right)\left(y-3\right)$.

Multiply the numerator and the denominator by the LCD.

$\frac{{\displaystyle \frac{y}{y+3}}}{2+{\displaystyle \frac{1}{y-3}}}=\frac{{\displaystyle \left(y+3\right)\left(y-3\right)\frac{y}{y+3}}}{\left(y+3\right)\left(y-3\right)\left(2+\frac{1}{y-3}\right)}$

Distribute the LCD in the denominator and then simplify.

$\frac{{\displaystyle \left(y+3\right)\left(y-3\right)\frac{y}{y+3}}}{\left(y+3\right)\left(y-3\right)\left(2+\frac{1}{y-3}\right)}=\frac{\left(y+3\right)\left(y-3\right)\frac{y}{y+3}}{\left(2\left(y+3\right)\left(y-3\right)+\left(y+3\right)\left(y-3\right).\frac{1}{y-3}\right)}$

Step 3. Now simplify the obtained expression.

$$\begin{gathered} \frac{\left(y+3\right)\left(y-3\right)\frac{y}{y+3}}{\left(2\left(y+3\right)\left(y-3\right)+\left(y+3\right)\left(y-3\right).\frac{1}{y-3}\right)}=\frac{\left(y-3\right)y}{\left(2\left(y+3\right)\left(y-3\right)+\left(y+3\right)\right)} \\ =\frac{y\left(y-3\right)}{2\left(y^2-9\right)+\left(y+3\right)} \\ =\frac{y\left(y-3\right)}{2y^2-18+y+3} \\ =\frac{y\left(y-3\right)}{2y^2+y-15} \\ =\frac{y^2-3y}{\left(y+3\right)\left(2y-5\right)} \end{gathered}$$