Chapter 5.5, Problem 65S

To determine

To solve the inequality:

  $6\left(5z-3\right)\le 36z$

And to graph the solution on a number line.

Answer to Problem 65S

  $\left[-3,\infty \right)$

  

Explanation of Solution

Given:

  $6\left(5z-3\right)\le 36z$

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:

To solve the inequality:

  $6\left(5z-3\right)\le 36z$

First, simplify this inequality as

  $\begin{array}{c}6\left(5z-3\right)\le 36z\\ \Rightarrow 30z-18\le 36z\end{array}$

Now, simplify further using properties of inequality, as

  $\begin{array}{c}30z-18\le 36z\\ \Rightarrow 30z-18-30z\le 36z-30z\\ \Rightarrow 0-18\le 6z\\ \Rightarrow -\frac{18}{6}\le \frac{6z}{6}\\ \Rightarrow -3\le z\end{array}$

Thus, the solution of given inequality is every number greater than and equals to $-3$ , this can be represented in interval form, as

  $\left[-3,\infty \right)$

Now, in order to sketch the graph of the solution of the inequality, since the interval of solution at left end point $-3$ is not strict, so use a solid circle at $-3$ . And the solution interval has $\infty$ at right endpoint which means every number greater than and equals to $-3$ is in the solution interval of given inequality. The graph of the solution of inequality is shown below: