Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Compound Inequalities
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Exercise 22 Page 204

Solve each inequality separately and compare the resulting solution sets.

Solution Set: z>2 or z<- 1
Graph:

Practice makes perfect

If we solve each inequality separately, we will find two solution sets. The union of those sets is the solution to the compound inequality.

First Inequality

We solve inequalities the same way we would solve equations. By adding 3 to both sides of the inequality, we can eliminate 3 from the left-hand side. This will help isolate z.

5z-3>7
5z>10
z>2

All values of z that are greater than 2 will satisfy the inequality. The inequality is strict meaning z cannot be equal to 2.

Second Inequality

By adding 6 to both sides of the inequality, we can begin to isolate z.

4z-6<- 10
4z<- 4
z<- 1

The second inequality is satisfied for all values of z less than - 1. The inequality is strict meaning z cannot be equal to - 1.

Comparing Solution Sets

Now we must compare our solution sets. From the first inequality, we know that the solution set consists of all numbers to the right of 2 on the number line, not including 2.

From the second inequality, we know the solution set included all values to the left of, but not including - 1.

The union of these solution sets is two intervals, z>2 or z<- 1.