Warm Up Solve. 1. y + 7 < –11 y < –18 2. 4m ≥ –12 m ≥ –3 3. 5 – 2x ≤ 17 x ≥ –6 Use interval notation to indicate the graphed numbers. 4. (-2, 3] (-, 1] 5.

Do Now 10/15/2015 Solve each equation. 1. |2v + 5| = 9 2 or –7 2. |5b| – 7 = 13 + 4

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

A conjunction is a compound statement that uses the word and. Conjunction: x ≥ –3 AND x < 2 Conjunctions can be written as a single statement as shown. x ≥ –3 and x< 2 –3 ≤ x < 2

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

Solve both inequalities for x. Example 1a Solve the compound inequality. Then graph the solution set and write the interval notation. x – 2 < 1 OR 5x ≥ 30 Solve both inequalities for x. x – 2 < 1 5x ≥ 30 or x ≥ 6 x < 3 –1 0 1 2 3 4 5 6 7 8 (–∞, 3) U [6, ∞)

Solve the compound inequality. Then graph the solution set. Example 1b Solve the compound inequality. Then graph the solution set. 2x ≥ –6 AND –x > –4 Solve both inequalities for x. 2x ≥ –6 and –x > –4 x ≥ –3 x < 4 –4 –3 –2 –1 0 1 2 3 4 5 [–3, 4)

Solve the compound inequality. Then graph the solution set. Try it out! Solve the compound inequality. Then graph the solution set. x –5 < 12 OR 6x ≤ 12 Solve both inequalities for x. x –5 < 12 or 6x ≤ 12 x < 17 x ≤ 2 2 4 6 8 10 12 14 16 18 20 (-∞, 17)

Solve the compound inequality. Then graph the solution set. Try It Out! 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 2 3 4 5 6 7 8 9 10 11 (4, 8]

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.

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

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.

Helpful Hint Think: Greator inequalities involving > or ≥ symbols are disjunctions. Think: Less thand inequalities involving < or ≤ symbols are conjunctions.

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

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

To check, you can test a point in each of the three region. 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. 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 (–∞, ∞)

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

Rewrite as compound inequality –3 ≤ x ≤ 13 Example 4a Continued Rewrite as compound inequality –3 ≤ x ≤ 13 –10 –5 0 5 10 15 20 25 [-3, 13]

Try it out! 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.