Chapter A.3, Problem 70E

Solution

Simplify the compound fraction.

  $\frac{1}{(b-a)}$ .

Given:

The given compound fraction is $\frac{\frac{1}{a}+\frac{1}{b}}{\frac{b}{a}-\frac{a}{b}}$ .

Concept Used:

• Compound fraction is a fraction with fractions in the numerator or denominator.
• First multiply the denominator of the fraction by the whole number. 
• Then, add that number to the numerator of the fraction, and keep the original denominator the same.

Calculation:

The given compound fraction is $\frac{\frac{1}{a}+\frac{1}{b}}{\frac{b}{a}-\frac{a}{b}}$ .

  $\begin{array}{c}\frac{\frac{1}{a}+\frac{1}{b}}{\frac{b}{a}-\frac{a}{b}}=\frac{\frac{b+a}{ab}}{\frac{b\cdot b-a\cdot a}{ab}}\\ =\frac{\frac{b+a}{ab}}{\frac{b^2-a^2}{ab}}\\ =\frac{ab(b+a)}{ab(b^2-a^2)}\\ =\frac{(b+a)}{(b^2-a^2)}\\ =\frac{(b+a)}{(b+a)(b-a)}\\ =\frac{1}{(b-a)}\end{array}$

Thus by simplifying the compound fraction $\frac{\frac{1}{a}+\frac{1}{b}}{\frac{b}{a}-\frac{a}{b}}$ the solution is $\frac{1}{(b-a)}$ .