Chapter 1.5, Problem 39PPS

To determine

To write and solve: the inequality given by the statement.

Answer to Problem 39PPS

An inequality is $\left(\frac{z}{3}+4\right)\le 2z+12$ .

Solution is $z\ge -\frac{24}{5}$ .

Explanation of Solution

Given information:

Given statement

The sum of one third of a number and 4 is at most the sum twice of the number and 12.

Calculation:

Consider the given statement,

The sum of one third of a number and 4 is at most the sum twice of the number and 12.

Let us consider a variable as $z$ to write an inequality

So, the inequality will be

  $\left(\frac{z}{3}+4\right)\le 2z+12$

Now, solve the inequality as shown,

  $\begin{array}{c}\left(\frac{z}{3}+4\right)\le 2z+12\\ \left(\frac{z}{3}+4-4-2z\right)\le 2z+12-4-2z\\ \frac{z}{3}-2z\le 8\\ \frac{-5z}{3}\le 8\\ z\ge -\frac{24}{5}\end{array}$

Hence, the solution of the inequality is $z\ge -\frac{24}{5}$ .