Chapter 6.6, Problem 35WE

To determine

Simplify: the given expression as fraction.

Answer to Problem 35WE

The simplified form of $\left(\frac{m}{n}-\frac{n}{m}\right)÷\left(\frac{1}{m}+\frac{1}{n}\right)$ is $m-n$.

Explanation of Solution

Given information:

The expression $\left(\frac{m}{n}-\frac{n}{m}\right)÷\left(\frac{1}{m}+\frac{1}{n}\right)$.

Formula used:

Least common denominator: It is the smallest multiple of denominator of set of fractions.

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$

The division of two fraction is product of dividend fraction and reciprocal of divisor fraction.

  $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

  $\left(\frac{m}{n}-\frac{n}{m}\right)÷\left(\frac{1}{m}+\frac{1}{n}\right)$

Recall that Least common denominator is the smallest multiple of denominator of set of fractions.

Least common denominator for both $\frac{m}{n},\frac{n}{m}$ and $\frac{1}{m},\frac{1}{n}$ is $mn$ . So,

  $=\left(\frac{m^2-n^2}{mn}\right)÷\left(\frac{n+m}{mn}\right)$

Recall that division of two fraction is done by $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$.

  $\begin{array}{l}=\left(\frac{m^2-n^2}{mn}\right)\times \left(\frac{mn}{m+n}\right)\\ =\left(\frac{m^2-n^2}{m+n}\right)\end{array}$

Factorizing the numerator of expression by using identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$

  $\begin{array}{l}=\frac{\left(m-n\right)\left(m+n\right)}{m+n}\\ =\boxed{m-n}\end{array}$

The simplified form of $\left(\frac{m}{n}-\frac{n}{m}\right)÷\left(\frac{1}{m}+\frac{1}{n}\right)$ is $m-n$.