## Algebra 1

Published by Prentice Hall

# Chapter 3 - Solving Inequalities - 3-6 Compound Inequalities - Practice and Problem-Solving Exercises - Page 204: 17

#### Answer

$b\lt-1$ or $b\gt2$

#### Work Step by Step

$6b-1\lt-7\longrightarrow$ add 1 to each side $6b-1+1\lt-7+1\longrightarrow$ add $6b\lt-6\longrightarrow$ divide each side by 6 $6b\div6\lt-6\div6\longrightarrow$ divide $b\lt-1$ OR $2b+1\gt5\longrightarrow$ subtract 1 from each side $2b+1-1\gt5-1\longrightarrow$ subtract $2b\gt4\longrightarrow$ divide each side by 2 $2b\div2\gt4\div2\longrightarrow$ divide $b\gt2$ Since -1 and 2 are not included in the solution sets, the endpoints on the graph are open circles.

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