Question

In the following exercise, Simplify.

$\frac{{\displaystyle \frac{2}{m}}+{\displaystyle \frac{m}{n}}}{{\displaystyle \frac{n}{m}}-{\displaystyle \frac{1}{n}}}$

Short answer

The solution is$\frac{2n+m^2}{n^2-m}$

Step-by-step solution

Step 1. Given information

$\frac{{\displaystyle \frac{2}{m}+\frac{m}{n}}}{{\displaystyle \frac{n}{m}-\frac{1}{n}}}$

Step 2. Simplify

The LCD of all the fractions is $nm$

Multiply both the numerator and denominator by $nm$

role="math" localid="1646724306783" $$\begin{gathered} =\frac{nm\left({\displaystyle \frac{2}{m}}+{\displaystyle \frac{m}{n}}\right)}{nm\left({\displaystyle \frac{n}{m}}-{\displaystyle \frac{1}{n}}\right)} \\ =\frac{{\displaystyle \frac{2n+m^2}{mn}}}{{\displaystyle \frac{n^2-m}{mn}}} \end{gathered}$$

Rewrite as fraction division.

role="math" localid="1646724387368" $=\frac{2n+m^2}{mn}÷\frac{n^2-m}{mn}$

Multiply the first the reciprocal of second

$$\begin{gathered} =\frac{2n+m^2}{mn}\times \frac{mn}{n^2-m} \\ =\frac{2n+m^2}{n^2-m} \end{gathered}$$