Question

In the following exercises, simplify:

$\frac{{\displaystyle \frac{2}{r}}+{\displaystyle \frac{2}{t}}}{{\displaystyle \frac{1}{r^2}}-{\displaystyle \frac{1}{t^2}}}$.

Short answer

The simplified rational expression is,$\frac{2rt}{t-r}$.

Step-by-step solution

Step 1. Given Information.

The rational expression is,

$\frac{{\displaystyle \frac{2}{r}}+{\displaystyle \frac{2}{t}}}{{\displaystyle \frac{1}{r^2}}-{\displaystyle \frac{1}{t^2}}}$.

Step 2. Simplifying the rational expression.

$\frac{{\displaystyle \frac{2}{r}}+{\displaystyle \frac{2}{t}}}{{\displaystyle \frac{1}{r^2}}-{\displaystyle \frac{1}{t^2}}}$

The LCD of the expression is $r^2{t}^2$.

$$\begin{gathered} =\frac{r^2{t}^2({\displaystyle \frac{2}{r}}+{\displaystyle \frac{2}{t}})}{r^2{t}^2({\displaystyle \frac{1}{r^2}}-{\displaystyle \frac{1}{t^2}})} \\ =\frac{r^2{t}^2\times {\displaystyle \frac{2}{r}}+r^2{t}^2\times {\displaystyle \frac{2}{t}}}{r^2{t}^2\times {\displaystyle \frac{1}{r^2}}-r^2{t}^2\times {\displaystyle \frac{1}{t^2}}} \\ =\frac{2r{t}^2+2r^{2}t}{t^2-r^2} \end{gathered}$$

Using $(a^2-b^2)=(a+b)(a-b)$in the denominator,

$=\frac{2rt(t+r)}{(t+r)(t-r)}$

Removing common factors,

$=\frac{2rt}{t-r}$