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Solving Absolute-Value Equations and Inequalities. 2-8. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 2. Holt Algebra 2. Warm Up Solve. 1. y + 7 < –11. y < –18. 2. 4 m ≥ –12. m ≥ –3. 3. 5 – 2 x ≤ 17. x ≥ –6.

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2-8

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

  2. 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.

  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. –6 –5 –4 –3 –2 –1 0 1 2 3 Example 1A: 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}. (–∞, –4) U [–2, ∞)

  9. and 2c + 1 < 1 c ≥ –4 c < 0 –6 –5 –4 –3 –2 –1 0 1 2 3 Example 1B: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. Solve both inequalities for c. The solution set is the set of points that satisfy both c ≥ –4 and c < 0. [–4, 0)

  10. –3 –2 –1 0 1 2 3 4 5 6 Example 1C: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. x – 5 < –2 OR –2x ≤ –10 Solve both inequalities for x. x – 5 < –2 or –2x ≤ –10 x < 3 x ≥ 5 The solution set is the set of all points that satisfy {x|x < 3 or x ≥ 5}. (–∞, 3) U [5, ∞)

  11. –1 0 1 2 3 4 5 6 7 8 Check It Out! Example 1a Solve the compound inequality. Then graph the solution set. x – 2 < 1OR 5x ≥ 30 Solve both inequalities for x. x– 2 < 1 5x ≥ 30 or x ≥ 6 x <3 The solution set is all points that satisfy {x|x < 3U x ≥6}. (–∞, 3) U [6, ∞)

  12. The solution set is the set of points that satisfy both {x|x ≥ –3 x < 4}. U –4 –3 –2 –1 0 1 2 3 4 5 Check It Out! 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 [–3, 4)

  13. 2 4 6 8 10 12 14 16 18 20 Check It Out! Example 1c 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 Because every point that satisfies x < 2 also satisfies x < 2, the solution set is {x|x < 17}. (-∞, 17)

  14. 2 3 4 5 6 7 8 9 10 11 Check It Out! Example 1d 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}. (4, 8]

  15. 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.

  16. 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.

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

  18. 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.

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

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

  21. 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

  22. Example 2B: Solving Absolute-Value Equations Solve the equation. Isolate the absolute-value expression. Rewrite the absolute value as a disjunction. Multiply both sides of each equation by 4. x = 16 or x = –16

  23. Check It Out! Example 2a Solve the equation. This can be read as “the distance from x to +9 is 4.” |x + 9| = 13 Rewrite the absolute value as a disjunction. x + 9 = 13 or x + 9 = –13 Subtract 9 from both sides of each equation. x = 4 or x = –22

  24. 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

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

  26. 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

  27. –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 

  28. –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 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. (–∞, ∞)

  29. 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

  30. –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 

  31. –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. (–∞, ∞)

  32. 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

  33. –6 –5 –3 –2–1 0 1 2 3 4 Example 4A Continued The solution set is {x|–5 ≤ x ≤2}.

  34. 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.

  35. 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

  36. –10 –5 0 5 10 15 20 25 Check It Out! Example 4 The solution set is {x|–3 ≤ x≤ 13}.

  37. 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.

  38. –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

  39. –4 –3 –2 –1 0 1 2 3 4 5 –4 –3 –2 –1 0 1 2 3 4 5 Lesson Quiz: Part II Solve. Then graph the solution. |1 – 2x| > 7 {x|x < –3 or x > 4} 5. |3k| + 11 > 8 6. R 7. –2|u + 7| ≥ 16 ø

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