1. Algebra 2 Lesson 1-6 Part 2 Absolute Value Inequalities

2. Goals Goal To write and solve inequalities involving absolute values. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

3. Essential Question Big Idea: Absolute Value Inequalities How can you solve an absolute value inequality? Students will understand that an absolute value quantity is nonnegative.

4. Vocabulary None

5. Absolute Value Inequalities When an inequality contains an absolute-value expression, it can be written as a compound inequality. The inequality |x| < 5 describes all real numbers whose distance from 0 is less than 5 units. The solutions are all numbers between –5 and 5, so |x| –5 AND x < 5.

6. The solutions of |x| < 3 are the points that are less than 3 units from zero. The solution is a conjunction: –3 < x < 3. “Less Than” - Conjunction

7. Think: Less thand inequalities involving < or ≤ symbols are conjunctions. Helpful Hint

8. “Less Than” Absolute Value Inequalities

9. Solve the inequality and graph the solutions. |x|– 3 < –1 x > –2 AND x < 2 Since 3 is subtracted from |x|, add 3 to both sides to undo the subtraction. +3 |x| < 2 |x|– 3 < –1 {x: –2 < x < 2}. Write as a compound inequality. The solution set is –2 –1012 2 units Example:

10. |x – 1| ≤ 2 Write as a compound inequality. Solve the inequality and graph the solutions. x – 1 ≥ –2 AND x – 1 ≤ 2 Solve each inequality. +1 x ≥ –1x ≤ 3AND {x: –1 ≤ x ≤ 3}. Write as a compound inequality. The solution set is –2–1 0 1 2 3 –3 Example:

11. Procedure Solving “Less Than” Inequalities Solving Absolute Inequalities: 1.Isolate the absolute value. 2.Break it into 2 inequalities (and statement) – one positive and the other negative reversing the sign. 3.Solve both inequalities. 4.Check both answers.

12. Just as you do when solving absolute-value equations, you first isolate the absolute-value expression when solving absolute-value inequalities. Helpful Hint

13. Solve the inequality and graph the solutions. 2|x| ≤ 6 x ≥ –3 AND x ≤ 3 Since x is multiplied by 2, divide both sides by 2 to undo the multiplication. |x| ≤ 3 {x: –3 ≤ x ≤ 3}. Write as a compound inequality. The solution set is 2|x| ≤ 6 2 –2 –1012 3 units –33 Your Turn:

14. |x + 3| – 4.5 ≤ 7.5 Since 4.5 is subtracted from |x + 3|, add 4.5 to both sides to undo the subtraction. Solve each inequality and graph the solutions. + 4.5 +4.5 |x + 3| ≤ 12 |x + 3| – 4.5 ≤ 7.5 x + 3 ≥ – 12 AND x + 3 ≤ 12 –3 x ≥ –15 AND x ≤ 9 Write as a compound inequality. {x: –15 ≤ x ≤ 9}. The solution set is –20 –15–10 –5 051015 Your Turn:

15. “Greater Than” Absolute Value Inequalities The inequality |x| > 5 describes all real numbers whose distance from 0 is greater than 5 units. The solutions are all numbers less than –5 or greater than 5. The inequality |x| > 5 can be rewritten as the compound inequality x 5.

16. The solutions of |x| > 3 are the points that are more than 3 units from zero. The solution is a disjunction: x 3. “Greater Than” - Disjunction

17. Think: Greator inequalities involving > or ≥ symbols are disjunctions. Helpful Hint

18. “Greater Than” Absolute Value Inequalities

19. Solve the inequality and graph the solutions. |x| + 14 ≥ 19 |x| ≥ 5 x ≤ –5 OR x ≥ 5 Since 14 is added to |x|, subtract 14 from both sides to undo the addition. Write as a compound inequality. The solution set is {x: x ≤ –5 OR x ≥ 5}. –10 –8 –6–4 –2 0246810 5 units – 14 –14 |x| + 14 ≥ 19 Example:

20. Solve the inequality and graph the solutions. 3 + |x + 2| > 5 Since 3 is added to |x + 2|, subtract 3 from both sides to undo the addition. Write as a compound inequality. Solve each inequality. |x + 2| > 2 – 3 – 3 3 + |x + 2| > 5 x + 2 2 –2 x 0 The solution set is {x: x 0}. Write as a compound inequality. –10 –8 –6–4 –2 02468 10 Example:

21. Procedure Solving “Greater Than” Inequalities Solving Absolute Inequalities: 1.Isolate the absolute value. 2.Break it into 2 inequalities (or statement) – one positive and the other negative reversing the sign. 3.Solve both inequalities. 4.Check both answers.

22. |x| + 10 ≥ 12 – 10 –10 |x| ≥ 2 –5 –4 –3–2 –1 012345 x ≤ –2 OR x ≥ 2 Write as a compound inequality. The solution set is {x: x ≤ –2 or x ≥ 2}. Since 10 is added to |x|, subtract 10 from both sides to undo the addition. Solve each inequality and graph the solutions. 2 units Your Turn:

23. Since is added to |x + 2 |, subtract from both sides to undo the addition. OR Solve the inequality and graph the solutions. x ≤ –6 x ≥ 1 Write as a compound inequality. The solution set is {x: x ≤ –6 or x ≥ 1} Write as a compound inequality. Solve each inequality. –7 –6 –5–4 –3 0123 –2–1 Your Turn:

24. A pediatrician recommends that a baby’s bath water be 95°F, but it is acceptable for the temperature to vary from this amount by as much as 3°F. Write and solve an absolute-value inequality to find the range of acceptable temperatures. Graph the solutions. Let t represent the actual water temperature. The difference between t and the ideal temperature is at most 3°F. t – 95 ≤ 3 Example: Application

25. t – 95 ≤ 3 |t – 95| ≤ 3 t – 95 ≥ –3 AND t – 95 ≤ 3 Solve the two inequalities. +95 t ≥ 92 AND t ≤ 98 The range of acceptable temperature is 92 ≤ t ≤ 98. 98 100 96 9492 90 Example: Continued

26. A dry-chemical fire extinguisher should be pressurized to 125 psi, but it is acceptable for the pressure to differ from this value by at most 75 psi. Write and solve an absolute-value inequality to find the range of acceptable pressures. Graph the solution. Let p represent the desired pressure. The difference between p and the ideal pressure is at most 75 psi. p – 125 ≤ 75 Your Turn:

27. p – 125 ≤ 75 |p – 125| ≤ 75 p – 125 ≥ –75 AND p – 125 ≤ 75 Solve the two inequalities. +125 p ≥ 50 AND p ≤ 200 The range of pressure is 50 ≤ p ≤ 200. 200225175150125100755025 Your Turn: Continued

28. Solutions to Absolute Value Inequalities When solving an absolute-value inequality, you may get a statement that is true for all values of the variable. In this case, all real numbers are solutions of the original inequality. If you get a false statement when solving an absolute- value inequality, the original inequality has no solutions. Its solution set is ø.

29. Solve the inequality. |x + 4|– 5 > – 8 + 5 |x + 4| > –3 Add 5 to both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers. The solution set is all real numbers. Example:

30. Solve the inequality. |x – 2| + 9 < 7 – 9 |x – 2| < –2 Subtract 9 from both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is false for all values of x. The inequality has no solutions. The solution set is ø. Example:

31. An absolute value represents a distance, and distance cannot be less than 0. Remember!

32. Solve the inequality. |x| – 9 ≥ –11 +9 ≥ +9 |x| ≥ –2 Add 9 to both sides. Absolute-value expressions are always nonnegative. Therefore, the statement is true for all real numbers. The solution set is all real numbers. Your Turn:

33. Solve the inequality. 4|x – 3.5| ≤ –8 44 |x – 3.5| ≤ –2 Absolute-value expressions are always nonnegative. Therefore, the statement is false for all values of x. Divide both sides by 4. The inequality has no solutions. The solution set is ø. Your Turn:

34. SUMMARY

35. Absolute Value Inequalities (<,  ) Inequalities of the Form < or  Involving Absolute Value If a is a positive real number and if u is an algebraic expression, then |u| -a |u|  a is equivalent to u  a and u > -a Note: If a = 0, |u| < 0 has no real solution, |u|  0 is equivalent to u = 0. If a < 0, the inequality has no real solution.

36. Absolute Value Inequalities (>,  ) Inequalities of the Form > or  Involving Absolute Value If a is a positive real number and u is an algebraic expression, then |u| > a is equivalent to u a |u|  a is equivalent to u  – a or u  a.

37. Procedure Solving Absolute Inequalities: 1.Isolate the absolute value. 2.Break it into 2 inequalites (“<, ≤ and statement” – “>, ≥ or statement”) – one positive and the other negative reversing the sign. 3.Solve both inequalities. 4.Check both answers.

38. Solving an Absolute-Value Equation Recall that  x  is the distance between x and 0. If  x  8, then any number between  8 and 8 is a solution of the inequality.  8  7  6  5  4  3  2  1 0 1 2 3 4 5 6 7 8 You can use the following properties to solve absolute-value inequalities and equations. Recall that | x | is the distance between x and 0. If | x |  8, then any number between  8 and 8 is a solution of the inequality.

39. S OLVING A BSOLUTE- V ALUE I NEQUALITIES a x  b  c and a x  b   c. a x  b  c and a x  b   c. a x  b  c or a x  b   c. a x  b  c or a x  b   c. | a x  b |  c | a x  b |  c | a x  b |  c | a x  b |  c means When an absolute value is less than a number, the inequalities are connected by and. When an absolute value is greater than a number, the inequalities are connected by or. SOLVING ABSOLUTE-VALUE INEQUALITIES

40. Solve | x  4 | < 3 Solving an Absolute-Value Inequality x  4 IS POSITIVE x  4 IS NEGATIVE | x  4 |  3 x  4   3 x  7 | x  4 |  3 x  4   3 x  1 Reverse inequality symbol.  This can be written as 1  x  7. The solution is all real numbers greater than 1 and less than 7. –10 –8 –6–4 –2 02468 10

41. 2x  1   9 | 2x  1 |  3  6 | 2x  1 |  9 2x   10 2x + 1 IS NEGATIVE  x   5 Solve | 2x  1 |  3  6 and graph the solution. | 2x  1 |  3  6 | 2x  1 |  9 2x  1  +9 2x  8 2x + 1 IS POSITIVE x  4 Solving an Absolute-Value Inequality Reverse inequality symbol. | 2x  1 |  3  6 | 2x  1 |  9 2x  1  +9 x  4 2x  8 | 2x  1 |  3  6 | 2x  1 |  9 2x  1   9 2x   10 x   5 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE  6  5  4  3  2  1 0 1 2 3 4 5 6 The solution is all real numbers greater than or equal to 4 or less than or equal to  5. This can be written as the compound inequality x   5 or x  4.  5 5 4.4.

42. MORE PROBLEMS FOR YOU

43. Solve the compound inequality. Then graph the solution set. |p – 2| ≤ –6 Multiply both sides by –2, and reverse the inequality symbol. Add 2 to both sides of each inequality. Rewrite the absolute value as a conjunction. |p – 2| ≤ –6 and p – 2 ≥ 6 p ≤ –4 and p ≥ 8 Because no real number satisfies both p ≤ –4 and p ≥ 8, there is no solution. The solution set is ø. Your Turn:

44. Solve the compound inequality. Then graph the solution set. |x – 5| ≤ 8 Multiply both sides by 2. Add 5 to both sides of each inequality. Rewrite the absolute value as a conjunction. x – 5 ≤ 8 and x – 5 ≥ –8 x ≤ 13 and x ≥ –3 Your Turn:

45. The solution set is {x|–3 ≤ x ≤ 13}. –10 –5 0 5 10 15 20 25 Continued

46. Solve the compound inequality. Then graph the solution set. |2x +7| ≤ 3 Multiply both sides by 3. Subtract 7 from both sides of each inequality. Divide both sides of each inequality by 2. Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 2x ≤ –4 and 2x ≥ –10 x ≤ –2 and x ≥ –5 Your Turn:

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

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

49. Solve: 1.|3k| + 11 > 8 2.–2|u + 7| ≥ 16 3.|1 – 2x| > 7 No Solution All Real Numbers x 4

50. Assignment Section 1-6 Part 2, Pg 49 – 52; #1 – 5 all, 6 – 52 even