Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Compound Inequalities
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Exercise 19 Page 204

Solve each inequality separately and compare the resulting solution sets.

Solution: d ≥ 2 or d<2
Graph:

Practice makes perfect

If we solve each inequality separately, we will find two solution sets. The union of those sets, the set containing values from one set or from the other, is the solution to the compound inequality.

First Inequality

Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.

4d+5 ≥ 13
4d ≥ 8
d ≥ 2

All values of d that are greater than or equal to 2 will satisfy the inequality.

Second Inequality

Again, we will solve the inequality by isolating the variable.

7d-2<12
7d<14
d<2

The second inequality is satisfied for all values of d less than 2. The inequality is strict meaning d cannot be equal to 2.

Comparing Solution Sets

Now we must compare our solution sets. From the first inequality, we know that the solution sets consist of all numbers to the right of 2 on the number line, including 2 itself.

From the second inequality, we know the solution sets included all values to the right of 2, but not including 2.

The union of these solution sets is two intervals d ≥ 2 or d<2. Notice that the union of these sets contains all real numbers.