Chapter 8.6, Problem 6AGP

To determine

To solve : the inequality

Answer to Problem 6AGP

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 2 Solve 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 4 Now 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 $0<x<\frac{93}{10}$ .