Chapter 5.3, Problem 38PPS

To determine

To solve the inequality:

  $8\left(a-2\right)\le 10\left(a+2\right)$

Answer to Problem 38PPS

  $\left[-18,\infty \right)$

Explanation of Solution

Given:

  $8\left(a-2\right)\le 10\left(a+2\right)$

Concept Used:

• 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.
• Properties of inequality:

1. $b\le c\Rightarrow b\pm a\le c\pm a$
2. $b\le c\Rightarrow ab\le ac,\text{ if }a>0$
3. $b\le c\Rightarrow ab\ge ac,\text{ if }a<0$

Calculation:

In order to solve the inequality

  $8\left(a-2\right)\le 10\left(a+2\right)$

First, simplify it’s both sides, as

  $\begin{array}{c}8\left(a-2\right)\le 10\left(a+2\right)\\ \Rightarrow 8a-16\le 10a+20\end{array}$

Now, subtract $8a$ from both sides of the inequality, as

  $\begin{array}{c}8a-16\le 10a+20\\ \Rightarrow 8a-16-8a\le 10a+20-8a\\ \Rightarrow 0-16\le 2a+20\\ \Rightarrow -16-20\le 2a+20-20\left(\begin{array}{l}\text{subtract 20 }\\ \text{from both sides}\end{array}\right)\\ \Rightarrow -36\le 2a\\ \Rightarrow -\frac{36}{2}\le \frac{2a}{2}\left(\begin{array}{l}\text{divide both}\\ \text{sides by 2 }\end{array}\right)\\ \Rightarrow -18\le a\end{array}$

Thus, the above inequality is true for every value of a greater than and equals to $-18$ , this can be represented in interval form, as

  $\left[-18,\infty \right)$