# Chapter 3 - Solving Inequalities - 3-6 Compound Inequalities - Practice and Problem-Solving Exercises - Page 204: 22

$z\gt2$ or $z\lt-1$

#### Work Step by Step

$5z-3\gt7\longrightarrow$ add 3 to each side $5z-3+3\gt7+3\longrightarrow$ add $5z\gt10\longrightarrow$ divide each side by 5 $5z\div5\gt10\div5\longrightarrow$ divide $z\gt2$ or $4z-6\lt-10$ add 6 to each side $4z-6+6\lt-10+6\longrightarrow$ add $4z\lt-4\longrightarrow$ divide each side by 4 $4z\div4\lt-4\div4\longrightarrow$ divide $z\lt-1$ Since 2 and -1 are not in the solution sets, the endpoint of the graph is an open circle.

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