Question

Solve each compound inequality. Then graph the solution set.

$\mathbf{3}\mathit{z}\mathbf{-}\mathbf{2}\mathbf{>}\mathbf{-}\mathbf{5}$and$\mathbf{7}\mathit{z}\mathbf{+}\mathbf{4}\mathbf{<}\mathbf{-}\mathbf{17}$

Short answer

The solution of the given compound inequality$3z-2>-5$ and$7z+4<-17$ is $\mathit{z}\mathbf{\in }\mathit{\varphi }$.

The graph of the solution set cannot be drawn as the solution of the compound inequality$3z-2>-5$ and$7z+4<-17$ is$\varphi$ that is an empty set.

Step-by-step solution

Step 1. Solve the given compound inequality 3z−2>−5 and 7z+4<−17.

The solution of the given compound inequality $3z-2>-5$and $7z+4<-17$is:

Solve the inequality $3z-2>-5$.

$\begin{array}{c}3z-2>-5\\ 3z-2+2>-5+2\\ 3z>-3\\ \frac{3z}{3}>\frac{-3}{3}\\ z>-1\\ z\in \left(-1,\infty \right)\end{array}$

Solve the inequality $7z+4<-17$.

$\begin{array}{c}7z+4<-17\\ 7z+4-4<-17-4\\ 7z<-21\\ \frac{7z}{7}<\frac{-21}{7}\\ z<-3\\ z\in \left(-\infty ,-3\right)\end{array}$

A compound inequality containing ‘and’ is true if both inequalities are true.

That implies the solution of the compound inequality containing ‘and’ is the intersection of the solutions of the two simple statements.

Find the intersection of the solutions of the inequalities$3z-2>-5$ and$7z+4<-17$ to find the solution of the compound inequality$3z-2>-5$ and $7z+4<-17$.

The intersection of the solutions of the inequalities $3z-2>-5$and $7z+4<-17$is:

$z\in \varphi$

Where$\varphi$ denotes the empty set.

Therefore, the solution of the compound inequality$3z-2>-5$ and$7z+4<-17$ is$\varphi$ that is an empty set.

Step 2. Draw the graph of the solution set.

The graph of the solution set cannot be drawn as the solution of the compound inequality$3z-2>-5$ and$7z+4<-17$ is that is an empty set.