Chapter 11.2, Problem 6P

To determine

Find the simplified form of the complex fraction: $\frac{\frac{1}{q+4}}{\frac{2q^2}{2q+8}}$

Answer to Problem 6P

The simplified form of the expression $\frac{\frac{1}{q+4}}{\frac{2q^2}{2q+8}}$ is $\frac{1}{q^2}$

Explanation of Solution

Given:

Find the simplified form of the complex fraction: $\frac{\frac{1}{q+4}}{\frac{2q^2}{2q+8}}$

Concept Used:

Find the factors of numerator and denominator, and then cancel the common factor and simplify.

For division of rational expression, we always flip the 2^nd expression and put a multiplication sign.

Rule: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}·\frac{d}{c}$

Calculation:

  $\begin{array}{l}\frac{\frac{1}{q+4}}{\frac{2q^2}{2q+8}}=\frac{1}{q+4}÷\frac{2q^2}{2q+8}\\ \\ \frac{1}{q+4}·\frac{2q+8}{2q^2}=\frac{1}{q+4}·\frac{2(q+4)}{2q^2}\\ \\ \frac{q+4}{q+4}·\frac{2}{2}·\frac{1}{q^2}=\frac{1}{q^2}\end{array}$

Thus, the simplified form of the expression $\frac{\frac{1}{q+4}}{\frac{2q^2}{2q+8}}$ is $\frac{1}{q^2}$