Chapter 8.6, Problem 14CYU

To determine

To calculate: The given inequality and check solution.

Answer to Problem 14CYU

the solution is $y<0$ or $y>\frac{52}{45}$

Explanation of Solution

Given information: Inequality $\frac{4}{3y}+\frac{2}{5y}<\frac{3}{2}$ .

Calculation:

Inequality $\frac{4}{3y}+\frac{2}{5y}<\frac{3}{2}$ .

The excluded value for this inequality is $0$ .

Solve the related equation.

  $\frac{4}{3y}+\frac{2}{5y}=\frac{3}{2}$

The LCD for the terms is $30y$ .

Multiply both side by LCD

  $\begin{array}{l}40+12=45y\\ 45y=52\\ y=\frac{52}{45}\end{array}$

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 $y=-1$

  $\begin{array}{l}-\frac{4}{3}-\frac{2}{5}<\frac{3}{2}\\ \frac{-26}{15}<\frac{3}{2}\end{array}$

Test $y=1$

  $\begin{array}{l}\frac{4}{3}+\frac{2}{5}<\frac{3}{2}\\ \frac{26}{15}<\frac{3}{2}\times \end{array}$

Test $y=2$

  $\begin{array}{l}\frac{4}{6}+\frac{2}{10}<\frac{3}{2}\\ \frac{52}{60}<\frac{3}{2}\end{array}$

The statement is true for $y=-1$ and $y=2$ . Therefore, the solution is $y<0$ or $y>\frac{52}{45}$ .