Chapter 5.6, Problem 37WE

To determine

To simplify the given expression.

Answer to Problem 37WE

  $\frac{2a}{\left(x+a\right)}$

Explanation of Solution

Given:

Expression: $\frac{x}{x-a}-\frac{x^2+a^2}{x^2-a^2}+\frac{a}{x+a}$

Calculation:

Consider the given expression $\frac{x}{x-a}-\frac{x^2+a^2}{x^2-a^2}+\frac{a}{x+a},$

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

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

  $\frac{x}{x-a}-\frac{x^2+a^2}{x^2-a^2}+\frac{a}{x+a}$

  $=\frac{x}{\left(x-a\right)}-\frac{x^2+a^2}{\left(x-a\right)\left(x+a\right)}+\frac{a}{\left(x+a\right)}$ [Factorizing the denominators]

  $=\frac{x\cdot \left(x+a\right)}{\left(x-a\right)\cdot \left(x+a\right)}-\frac{x^2+a^2}{\left(x-a\right)\left(x+a\right)}+\frac{a\cdot \left(x-a\right)}{\left(x+a\right)\cdot \left(x-a\right)}$

  $=\frac{x^2+ax}{\left(x-a\right)\left(x+a\right)}-\frac{x^2+a^2}{\left(x-a\right)\left(x+a\right)}+\frac{ax-a^2}{\left(x-a\right)\left(x+a\right)}$

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

  $=\frac{x^2+ax-\left(x^2+a^2\right)+\left(ax-a^2\right)}{\left(x-a\right)\left(x+a\right)}$

  $\begin{array}{l}=\frac{x^2+ax-x^2-a^2+ax-a^2}{\left(x-a\right)\left(x+a\right)}\\ =\frac{2ax-2a^2}{\left(x-a\right)\left(x+a\right)}\\ =\frac{2a\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\end{array}$

  $=\frac{2a}{\left(x+a\right)}$

Conclusion:

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