Chapter 6.6, Problem 33WE

To determine

Simplify: the given expression as fraction.

Answer to Problem 33WE

The simplified form of $\left(\frac{a+b}{a}-1\right)\left(\frac{a}{b}+1\right)$ is $\frac{\left(2x^2+3\right)\left(x^2-2\right)}{x^2}$.

Explanation of Solution

Given information:

The expression $\left(\frac{a+b}{a}-1\right)\left(\frac{a}{b}+1\right)$.

Formula used:

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

  $\begin{array}{l}\left(\frac{a+b}{a}-1\right)\left(\frac{a}{b}+1\right)\\ =\frac{a+b}{a}\cdot \frac{a}{b}+\frac{a+b}{a}\cdot 1-1\cdot \frac{a}{b}-1\cdot 1\\ =\frac{a+b}{b}+\frac{a+b}{a}-\frac{a}{b}-1\end{array}$

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

  $=\frac{a+b}{b}+\frac{a+b}{a}-\frac{a}{b}-\frac{1}{1}$

Least common denominator for $\frac{a+b}{b},\frac{a+b}{a},\frac{a}{b},\frac{1}{1}$ is $ab$ . So,

  $\begin{array}{l}=\frac{a\left(a+b\right)+b\left(a+b\right)-a\cdot b-a\cdot b}{a}\\ =\frac{a^2+ab+ab+b^2-ab-ab}{a}\\ =\boxed{\frac{a^2+b^2}{a}}\end{array}$

The simplified form of $\left(\frac{a+b}{a}-1\right)\left(\frac{a}{b}+1\right)$ is $\frac{a^2+b^2}{a}$.