A compound statement is made up of more than one equation or inequality. A disjunction is a compound statement that uses the word or. Disjunction: x ≤ –3 OR x > 2 Set builder notation: {x|x ≤ –3 U x > 2} A disjunction is true if and only if at least one of its parts is true.

A conjunction is a compound statement that uses the word and. Conjunction: x ≥ –3 AND x < 2 Set builder notation: {x|x ≥ –3 x < 2}. A conjunction is true if and only if all of its parts are true. Conjunctions can be written as a single statement as shown. x ≥ –3 and x< 2 –3 ≤ x < 2 U

Dis- means “apart. ” Disjunctions have two separate pieces Dis- means “apart.” Disjunctions have two separate pieces. Con- means “together” Conjunctions represent one piece. Reading Math

Example 1: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. 6y < –24 OR y +5 ≥ 3 Solve both inequalities for y. 6y < –24 y + 5 ≥3 or y < –4 y ≥ –2 The solution set is all points that satisfy {y|y < –4 or y ≥ –2}. –6 –5 –4 –3 –2 –1 0 1 2 3 (–∞, –4) U [–2, ∞)

Solve the compound inequality. Then graph the solution set. Example 2 Solve the compound inequality. Then graph the solution set. –3x < –12 AND x + 4 ≤ 12 Solve both inequalities for x. –3x < –12 and x + 4 ≤ 12 x > 4 x ≤ 8 The solution set is the set of points that satisfy both {x|4 < x ≤ 8}. 2 3 4 5 6 7 8 9 10 11 (4, 8]

Example 3: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. X +3 < 7 OR 3x > 18 Solve both inequalities for x. x + 3 < 7 or 3x > 18 x < 4 x > 6 The solution set is the set of all points that satisfy {x|x < 4 or x > 6}. –3 –2 –1 0 1 2 3 4 5 6 (–∞, 4) U (6, ∞)

Recall that the absolute value of a number x, written |x|, is the distance from x to zero on the number line. Because absolute value represents distance without regard to direction, the absolute value of any real number is nonnegative.

Absolute-value equations and inequalities can be represented by compound statements. Consider the equation |x| = 3. The solutions of |x| = 3 are the two points that are 3 units from zero. The solution is a disjunction: x = –3 or x = 3.

Example 2A: Solving Absolute-Value Equations Solve the equation. This can be read as “the distance from k to –3 is 10.” |–3 + k| = 10 Rewrite the absolute value as a disjunction. –3 + k = 10 or –3 + k = –10 Add 3 to both sides of each equation. k = 13 or k = –7

Check It Out! Example 2b Solve the equation. |6x| – 8 = 22 Isolate the absolute-value expression. |6x| = 30 Rewrite the absolute value as a disjunction. 6x = 30 or 6x = –30 Divide both sides of each equation by 6. x = 5 or x = –5

You can solve absolute-value inequalities using the same methods that are used to solve an absolute-value equation.

Note: The symbol ≤ can replace <, and the rules still apply Note: The symbol ≤ can replace <, and the rules still apply. The symbol ≥ can replace >, and the rules still apply.

The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3.

Check It Out! Example 4a Solve the compound inequality. Then graph the solution set. |x – 5| ≤ 8 Multiply both sides by 2. Rewrite the absolute value as a conjunction. x – 5 ≤ 8 and x – 5 ≥ –8 Add 5 to both sides of each inequality. x ≤ 13 and x ≥ –3

The solution set is {x|–3 ≤ x ≤ 13}. Check It Out! Example 4 The solution set is {x|–3 ≤ x ≤ 13}. –10 –5 0 5 10 15 20 25

The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x < –3 or x > 3.

Example 3A: Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |–4q + 2| ≥ 10 Rewrite the absolute value as a disjunction. –4q + 2 ≥ 10 or –4q + 2 ≤ –10 Subtract 2 from both sides of each inequality. –4q ≥ 8 or –4q ≤ –12 Divide both sides of each inequality by –4 and reverse the inequality symbols. q ≤ –2 or q ≥ 3

To check, you can test a point in each of the three region. Example 3A Continued {q|q ≤ –2 or q ≥ 3} –3 –2 –1 0 1 2 3 4 5 6 (–∞, –2] U [3, ∞) To check, you can test a point in each of the three region. |–4(–3) + 2| ≥ 10 |14| ≥ 10  |–4(0) + 2| ≥ 10 |2| ≥ 10 x |–4(4) + 2| ≥ 10 |–14| ≥ 10 

Check It Out! Example 3a Solve the inequality. Then graph the solution. |4x – 8| > 12 Rewrite the absolute value as a disjunction. 4x – 8 > 12 or 4x – 8 < –12 Add 8 to both sides of each inequality. 4x > 20 or 4x < –4 Divide both sides of each inequality by 4. x > 5 or x < –1

Check It Out! Example 3a Continued {x|x < –1 or x > 5} –3 –2 –1 0 1 2 3 4 5 6 (–∞, –1) U (5, ∞) To check, you can test a point in each of the three region. |4(–2) + 8| > 12 |–16| > 12  |4(0) + 8| > 12 |8| > 12 x |4(6) + 8| > 12 |32| > 12 

Solve the inequality. Then graph the solution. Check It Out! Example 3b Solve the inequality. Then graph the solution. |3x| + 36 > 12 Isolate the absolute value as a disjunction. |3x| > –24 Rewrite the absolute value as a disjunction. 3x > –24 or 3x < 24 Divide both sides of each inequality by 3. x > –8 or x < 8 The solution is all real numbers, R. –3 –2 –1 0 1 2 3 4 5 6 (–∞, ∞)

Check It Out! Example 4b Solve the compound inequality. Then graph the solution set. –2|x +5| > 10 Divide both sides by –2, and reverse the inequality symbol. |x + 5| < –5 Rewrite the absolute value as a conjunction. x + 5 < –5 and x + 5 > 5 Subtract 5 from both sides of each inequality. x < –10 and x > 0 Because no real number satisfies both x < –10 and x > 0, there is no solution. The solution set is ø.