Chapter 9.6, Problem 6ACYP

To determine

To solve : the inequality

Answer to Problem 6ACYP

The solution of the inequality is $0<x<\frac{93}{10}$

Explanation of Solution

Given information :

The inequality is

  $\frac{5}{x}+\frac{6}{5x}>\frac{2}{3}$

Calculation :

The solution of the inequality is

  $\frac{5}{x}+\frac{6}{5x}>\frac{2}{3}$

Step 1 The excluded value for this inequality is 1.

Step 2Solve the related equation

  $\begin{array}{l}\frac{5}{x}+\frac{6}{5x}=\frac{2}{3}\\ \frac{25+6}{5x}=\frac{2}{3}\\ \frac{31}{5x}=\frac{2}{3}\\ 10x=93\\ x=\frac{93}{10}\\ x=9.3\end{array}$

Step 3 Draw vertical lines at the excluded value and at the solution to separate the number line into intervals.

Step 4Now test sample value in each interval to determine whether the value to the interval satisfy the inequality.

  $\begin{array}{l}\text{Test x}=8\\ \frac{5}{x}+\frac{6}{5x}>\frac{2}{3}\\ \frac{5}{8}+\frac{6}{5\cdot 8}>\frac{2}{3}\\ \frac{5}{8}+\frac{6}{40}>\frac{2}{3}\\ \frac{5\cdot 5+6\cdot 1}{40}>\frac{2}{3}\\ \frac{25+6}{40}>\frac{2}{3}\\ \frac{31}{40}>\frac{2}{3}\\ 0.775>0.67\end{array}$

  $\begin{array}{l}\text{Test x}=10\\ \frac{5}{x}+\frac{6}{5x}>\frac{2}{3}\\ \frac{5}{10}+\frac{6}{5\cdot 10}>\frac{2}{3}\\ \frac{1}{2}+\frac{3}{25}>\frac{2}{3}\\ \frac{25\cdot 1+2\cdot 3}{50}>\frac{2}{3}\\ \frac{25+6}{50}>\frac{2}{3}\\ \frac{31}{50}>\frac{2}{3}\\ \frac{31}{50}<\frac{2}{3}\end{array}$

  $\begin{array}{l}\text{Test x}=-2\\ \frac{5}{x}+\frac{6}{5x}>\frac{2}{3}\\ \frac{5}{-2}+\frac{6}{5\cdot -2}>\frac{2}{3}\\ \frac{-5}{2}-\frac{6}{10}>\frac{2}{3}\\ \frac{-50-12}{20}>\frac{2}{3}\\ \frac{-62}{20}>\frac{2}{3}\\ \frac{-31}{10}>\frac{2}{3}\\ \frac{10}{31}<\frac{-3}{2}\end{array}$

  $\begin{array}{l}\text{Test x}=2\\ \frac{5}{2}+\frac{6}{5\cdot 2}>\frac{2}{3}\\ \frac{5}{2}+\frac{6}{10}>\frac{2}{3}\\ \frac{25+6}{10}>\frac{2}{3}\\ \frac{31}{10}>\frac{2}{3}\\ 3.1>0.67\end{array}$

The statement is true for $x=2\text{ and }8$

Hence, the solution of the inequality is $0<x<\frac{93}{10}$ .