Solving Absolute Value Equations and Inequalities

1. Solving Absolute-Value Equations and Inequalities 2-8 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

2. –4 –3 –2 –1 0 1 2 3 4 5 –4 –3 –2 –1 0 1 2 3 4 5 Lesson Quiz: Part I Solve. Then graph the solution. 1. y – 4 ≤ –6 or 2y >8 {y|y ≤ –2 ≤ or y > 4} {x|–3 < x ≤ –1} 2. –7x < 21 and x + 7 ≤ 6 Solve each equation. 3. |2v + 5| = 9 4. |5b| – 7 = 13 2 or –7 + 4

3. Objectives Solve compound inequalities. Write and solve absolute-value equations and inequalities.

4. Vocabulary disjunction conjunction absolute-value

5. 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 ORx > 2 Set builder notation: {x|x ≤ –3 Ux > 2} A disjunction is true if and only if at least one of its parts is true.

6. Conjunction: x ≥ –3 ANDx < 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 A conjunction is a compound statement that uses the word and.

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

8. 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.

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

10. 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.

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

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

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

14. 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

15. –3 –2 –1 0 1 2 3 4 5 6 Example 3A Continued {q|q ≤ –2 or q ≥ 3} (–∞, –2] U [3, ∞) To check, you can test a point in each of the three region. |–4(0) + 2| ≥ 10 |2| ≥ 10 x |–4(–3) + 2| ≥ 10 |14| ≥ 10  |–4(4) + 2| ≥ 10 |–14| ≥ 10 

16. –3 –2 –1 0 1 2 3 4 5 6 Example 3B: Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |0.5r| – 3 ≥ –3 (always true?? – all reals) Isolate the absolute value as a disjunction. |0.5r| ≥ 0 Rewrite the absolute value as a disjunction. 0.5r ≥ 0 or 0.5r ≤ 0 Divide both sides of each inequality by 0.5. r ≤0 or r ≥ 0 The solution is allreal numbers, R. (–∞, ∞)

17. 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

18. –3 –2 –1 0 1 2 3 4 5 6 Check It Out! Example 3a Continued {x|x < –1 or x > 5} (–∞, –1) U (5, ∞) To check, you can test a point in each of the three region. |4(0) + 8| > 12 |8| > 12 x |4(–2) + 8| > 12 |–16| > 12  |4(6) + 8| > 12 |32| > 12 

19. –3 –2 –1 0 1 2 3 4 5 6 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 allreal numbers, R. (–∞, ∞)

20. Example 4A: Solving Absolute-Value Inequalities with Conjunctions Solve the compound inequality. Then graph the solution set. Multiply both sides by 3. |2x +7| ≤ 3 Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 Subtract 7 from both sides of each inequality. 2x ≤ –4 and 2x ≥ –10 Divide both sides of each inequality by 2. x ≤ –2 and x ≥ –5

21. –6 –5 –3 –2–1 0 1 2 3 4 Example 4A Continued The solution set is {x|–5 ≤ x ≤2}.

22. Example 4B: Solving Absolute-Value Inequalities with Conjunctions Solve the compound inequality. Then graph the solution set. Multiply both sides by –2, and reverse the inequality symbol. |p – 2| ≤ –6 Rewrite the absolute value as a conjunction. |p – 2| ≤ –6 and p – 2 ≥ 6 Add 2 to both sides of each inequality. p ≤ –4 and p ≥ 8 Because no real number satisfies both p ≤ –4 andp ≥ 8, there is no solution. The solution set is ø.

23. Check It Out! Example 4a Solve the compound inequality. Then graph the solution set. Multiply both sides by 2. |x – 5| ≤ 8 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

24. –10 –5 0 5 10 15 20 25 Check It Out! Example 4 The solution set is {x|–3 ≤ x≤ 13}.

25. Check It Out! Example 4b Solve the compound inequality. Then graph the solution set. –2|x +5| > 10 (Never True??-no sol.) Divide both sides by –2, and revrse 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.

26. Assignment Page 154 # 8-13 all, 20-36 extra credit if you want it Due Wednesday Quiz Tomorrow