Chapter 5.6, Problem 31WE

To determine

To simplify the given expression.

Answer to Problem 31WE

  $\frac{2y}{\left(x-y\right)\left(x+y\right)}$

Explanation of Solution

Given:

Expression: ${\left(x-y\right)}^{-1}-{\left(x+y\right)}^{-1}$

Calculation:

Consider the given expression ${\left(x-y\right)}^{-1}-{\left(x+y\right)}^{-1}$ ,

It can also be written as $\frac{1}{x-y}-\frac{1}{x+y}$ .

Since the denominators of both the fractions are different, find the least common denominator for both the fractions.

Here, least common denominator is $\left(x-y\right)\left(x+y\right)$ , so

  $\frac{1}{x-y}-\frac{1}{x+y}$

  $=\frac{1\cdot \left(x+y\right)}{\left(x-y\right)\cdot \left(x+y\right)}-\frac{1\cdot \left(x-y\right)}{\left(x+y\right)\cdot \left(x-y\right)}$

  $=\frac{\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}-\frac{\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}$

Now, as the denominators of the two fractions are same, simply subtract the numerators and put the result on the same denominator.

  $=\frac{\left(x+y\right)-\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}$

  $\begin{array}{l}=\frac{x+y-x+y}{\left(x-y\right)\left(x+y\right)}\\ =\frac{\cancel{x}+y-\cancel{x}+y}{\left(x-y\right)\left(x+y\right)}\\ =\frac{2y}{\left(x-y\right)\left(x+y\right)}\end{array}$

Conclusion:

Therefore, after simplification, the answer is $\frac{2y}{\left(x-y\right)\left(x+y\right)}$ .