Chapter 9.6, Problem 12CYU

To determine

To calculate: The given inequality and check solution.

Answer to Problem 12CYU

the solution is $0<x<\frac{23}{20}$ .

Explanation of Solution

Given information: Inequality $\frac{3}{5x}+\frac{1}{6x}>\frac{2}{3}$ .

Calculation:

Inequality $\frac{3}{5x}+\frac{1}{6x}>\frac{2}{3}$ .

The excluded value for this inequality is $0$ .

Solve the related equation.

  $\frac{3}{5x}+\frac{1}{6x}=\frac{2}{3}$

The LCD for the terms is $30x$ .

Multiply both side by LCD

  $\begin{array}{l}18+5=20x\\ 20x=23\\ x=\frac{23}{20}\end{array}$

Draw vertical lines at the excluded value and at the solutions to separates the number line into intervals.

  

Now test a sample value in each interval to determine whether the values in the interval satisfy the inequality.

Test $x=-1$

  $\begin{array}{l}-\frac{3}{5}-\frac{1}{6}>\frac{2}{3}\\ \frac{-23}{30}>\frac{2}{3}\times \end{array}$

Test $x=1$

  $\begin{array}{l}\frac{3}{5}+\frac{1}{6}>\frac{2}{3}\\ \frac{23}{30}>\frac{2}{3}\end{array}$

Test $x=2$

  $\begin{array}{l}\frac{3}{10}+\frac{1}{12}>\frac{2}{3}\\ \frac{23}{30}>\frac{2}{3}\times \end{array}$

The statement is true for $x=1$ . Therefore, the solution is $0<x<\frac{23}{20}$ .