Chapter 2.1, Problem 23E

To determine

Whether z = 5 is the solution of an inequality $20\le \frac{10}{2z}+20$ .

Answer to Problem 23E

z = 5 is the solution of an inequality $20\le \frac{10}{2z}+20$ .

Explanation of Solution

Given:

The inequality is $20\le \frac{10}{2z}+20$ .

Calculation:

The inequality is in the ratio. Determine the zeroes of numerator and denominator for the critical number as follows:

Simplify the inequality as follows:

  $\begin{array}{l}20\le \frac{10}{2z}+20\\ \frac{10}{2z}+20-20\ge 20-20\\ \frac{10}{2z}\ge 0\end{array}$ For denominator,

  $\begin{array}{c}2z=0\\ z=0\end{array}$

The value of z cannot be zero.

Take this critical point on the number line as shown below:

  

Substitute the values from three regions in the inequality to satisfy the region.

For z = 1,

  $\begin{array}{l}20\le \frac{10}{2\left(1\right)}+20\\ 20\le 25\end{array}$

This satisfies the inequality. Thus, the region is $\left(0,\infty \right)$ .

  

Hence, z can be 5 as it is in the interval $\left(0,\infty \right)$ .