Chapter 1.5, Problem 47PPSE

Solution

To solve the inequality $\frac{2}{3}\left(x-12\right)\le x+8$ and graph the solution.

  $x\ge -48$

The graph of solution set is shown below:

  

Given:

The compound inequality, $\frac{2}{3}\left(x-12\right)\le x+8$ .

Concept Used:

• If there is the strict inequality symbol $<\text{ or}>$ , then use a parenthesis $\left(\right)$ on the interval
• If there is the inequality symbol $\le \text{ or}\ge$ , then use a bracket $\left[\right]$ on the interval
• To graph the endpoint with strict inequality $<\text{ or}>$ , use the open parenthesis or a hollow circle at that endpoint.
• To graph the endpoint with inequality symbol $\le \text{ or}\ge$ , use the bracket or a solid circle at that endpoint.
• Multiplying or dividing an inequality by a negative number reverse the inequality sign.

Calculation:

In order to solve the given inequality $\frac{2}{3}\left(x-12\right)\le x+8$ , perform the basic algebraic operations to isolate the variable term as shown below

  $\begin{array}{c}\frac{2}{3}\left(x-12\right)\le x+8\\ \frac{2}{3}x-8\le x+8\\ \frac{2}{3}x-x-8\le x-x+8\\ \frac{2x-3x}{3}-8\le 8\\ \frac{-x}{3}-8+8\le 8+8\\ \frac{-x}{3}\le 16\\ \frac{-x}{3}\cdot \left(-3\right)\ge 16\cdot \left(-3\right)\\ x\ge -48\end{array}$

Thus, the solution of the given inequality is $x\ge -48$ .

Since the inequality sign is not strict, use the solid circles at $-48$ and shade the region to the right side as shown below