COMPOUND INEQUALITIES AND OR

Compound Inequalities with "And"

A compound inequality containing and is true only if both inequalities are true. Thus, the graph of a compound inequality containing and is the intersection of the graphs of the two inequalities.

In other words, the solution must be a solution of both inequalities.

That is, we have to shade the portion that we find in both graphs. If the particular region is not in one of the graphs, we should not shade the region.

Compound Inequalities with "Or"

Another type of compound inequality contains the word or. A compound inequality containing or is true if one or more of the inequalities is true. The graph of a compound inequality containing or is the of the graphs of the two inequalities.

In other words, the solution of the compound inequality is a solution of either inequality, not necessarily both. The union can be found by graphing each inequality.

Example 1 :

Solve each compound inequality. Then graph the solution set.

h - 10 < - 21  (or)  h + 3 < 2

Solution : 

h - 10 < - 21  (or)  h + 3 < 2

Now let us graph for the first inequality

h < -11

We have to shade the portion, which is lesser than -11. Since it is less than sign, we have to use the unfilled circle 

Now let us graph for the second inequality

h < -1

We have to shade the portion, which is lesser than -1. Since it is less than sign, we have to use the unfilled circle 

By combining the above two graphs, we get

Example 2 :

Solve the compound inequality. Then graph the solution set

k + 2 > 12 and k + 2 ≤ 18

Solution :

By graphing the inequality k > 10, we get the graph given below.

By graphing the inequality ≤ 16, we get the graph given below.

By combining the above two graphs, we get the common region between 10 and 16.

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