Chapter 5.6, Problem 45PPS

To determine

To solve the compound inequality:

  $-4a+13\ge 29,\text{ and }10<6a-14$

And to graph the solution on a number line.

Answer to Problem 45PPS

No solution.

Explanation of Solution

Given:

  $-4a+13\ge 29,\text{ and }10<6a-14$

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 compound inequality:

  $-4a+13\ge 29,\text{ and }10<6a-14$

First, solve the inequality

  $-4a+13\ge 29$

First, subtract 13 from both sides, as

  $\begin{array}{c}-4a+13-13\ge 29-13\\ \Rightarrow -4a\ge 16\end{array}$

Now, divide both sides by $-4$ , that would change the inequality, as

  $\begin{array}{c}-4a\ge 16\\ \Rightarrow -\frac{4a}{-4}\le \frac{16}{-4}\\ \Rightarrow a\le -4\end{array}$

Now, solve the inequality

  $10<6a-14$

First, by add 14 both sides, as

  $\begin{array}{c}10+14<6a-14+14\\ \Rightarrow 24\le 6a\end{array}$

Now, divide both sides of above inequality by 6, as

  $\begin{array}{c}24\le 6a\\ \Rightarrow \frac{24}{6}\le \frac{6a}{6}\\ \Rightarrow 4\le a\end{array}$

Combining both of solutions:

  $4\le a\text{ and }a\le -4$

But there does not exist any number a which is less than $-4$ but greater than 4.

Thus, the given compound inequality has no solution.