Warm Up

1. Preview Warm Up California Standards Lesson Presentation

2. x < –3 –3 –2 0 1 2 3 4 5 –5 –4 –1 x ≥ 2 –3 –2 0 1 2 3 4 5 –5 –4 –1 x > –2 –3 –2 0 1 2 3 4 5 –5 –4 –1 Warm Up Solve each inequality and graph the solution. 1. x + 7 < 4 2. 14x ≥ 28 3. 5 + 2x > 1

3. California Standards 3.0 Students solve equations and inequalities involving absolute values. Also covered: 5.0

4. 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 can be rewritten as –5 < x < 5, or as x > –5 AND x < 5.

6. |x|– 3 < –1 +3 +3 |x| < 2 Write as a compound inequality. {x: –2 < x < 2}. The solution set is 2 units 2 units –1 0 1 2 –2 Additional Example 1A: Solving Absolute-Value Inequalities Involving < Solve the inequality and graph the solutions. |x|– 3 < –1 Since 3 is subtracted from |x|, add 3 to both sides to undo the subtraction. x > –2 AND x < 2

7. +1 +1 +1 +1 x ≥ –1 AND x ≤ 3 Write as a compound inequality. The solution set is {x: –1 ≤ x ≤ 3}. 0 –3 –2 –1 1 2 3 Additional Example 1B: Solving Absolute-Value Inequalities Involving < Solve the inequality and graph the solutions. |x – 1| ≤ 2 x – 1 ≥ –2 AND x – 1 ≤ 2 Write as a compound inequality. Solve each inequality.

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

9. 2|x| ≤ 6 2 2 Write as a compound inequality. {x: –3 ≤ x ≤ 3}. The solution set is 3 units 3 units –1 0 1 2 –3 3 –2 Check It Out! Example 1a Solve the inequality and graph the solutions. 2|x| ≤ 6 Since x is multiplied by 2, divide both sides by 2 to undo the multiplication. |x| ≤ 3 x ≥ –3 AND x ≤ 3

10. |x + 3|– 4.5≤ 7.5 +4.5 +4.5 |x + 3| ≤ 12 –3 –3 –3 –3 The solution set is x ≥ –15 AND x ≤ 9 {x: –15 ≤ x ≤ 9}. –15 –10 0 5 10 15 –20 –5 Check It Out! Example 1b Solve each inequality and graph the solutions. |x + 3|– 4.5≤ 7.5 Since 4.5 is subtracted from |x + 3|, add 4.5 to both sides to undo the subtraction. x + 3 ≥ –12 AND x + 3 ≤ 12 Write as a compound inequality.

11. 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 OR x > 5.

13. |x| + 14 ≥ 19 – 14 –14 5 units 5 units –8 –2 –10 –6 –4 0 2 4 6 8 10 Additional Example 2A: Solving Absolute-Value Inequalities Involving > Solve the inequality and graph the solutions. |x| + 14 ≥ 19 Since 14 is added to |x|, subtract 14 from both sides to undo the addition. |x| ≥ 5 Write as a compound inequality. The solution set is {x: x ≤ –5 OR x ≥ 5}. x ≤ –5 OR x ≥ 5

14. 3 + |x + 2| > 5 – 3 – 3 |x + 2| > 2 x + 2 < –2 OR x + 2 > 2 –2 –2 –2 –2 x < –4 OR x > 0 10 –8 –2 –10 –6 –4 0 2 4 6 8 Additional Example 2B: Solving Absolute-Value Inequalities Involving > 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. Write as a compound inequality. The solution set is {x: x < –4 or x > 0}.

15. – 10 –10 |x| ≥2 2 units 2 units –3 –2 0 1 2 3 4 5 –5 –4 –1 Check It Out! Example 2a Solve each inequality and graph the solutions. |x| + 10 ≥ 12 |x| + 10 ≥ 12 Since 10 is added to |x|, subtract 10 from both sides to undo the addition. x ≤ –2 OR x ≥ 2 Write as a compound inequality. The solution set is {x: x ≤ –2 or x ≥ 2}.

16. Since is added to |x + 2 |, subtract from both sides to undo the addition. x ≤ –6 x ≥ 1 OR Check It Out! Example 2b Solve the inequality and graph the solutions. Write as a compound inequality. Solve each inequality. Write as a compound inequality. The solution set is {x: x ≤ –6 or x ≥ 1}

17. –5 –4 0 1 2 3 –2 –1 –7 –6 –3 Check It Out! Example 2b Continued Solve the inequality and graph the solutions.

18. Additional Example 3: Application 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

19. +95 +95 +95 +95 t ≥ 92 AND t ≤ 98 92 94 90 96 98 100 Additional Example 3 Continued t – 95 ≤ 3 |t – 95| ≤ 3 Solve the two inequalities. t – 95 ≥ –3 AND t – 95 ≤ 3 The range of acceptable temperature is 92 ≤ t ≤ 98.

20. Check It Out! Example 3 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

21. +125 +125 +125 +125 p ≥ 50 AND p ≤ 200 25 50 75 100 125 150 175 200 225 Check It Out! Example 3 Continued p – 125 ≤ 75 |p – 125| ≤ 75 Solve the two inequalities. p – 125 ≥ –75 AND p – 125 ≤ 75 The range of pressure is 50 ≤ p ≤ 200.

22. 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.

23. |x + 4|– 5 > – 8 + 5 + 5 |x + 4| > –3 Additional Example 4A: Special Cases of Absolute-Value Inequalities Solve the inequality. |x + 4|– 5 > – 8 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.

24. |x – 2| + 9 < 7 – 9 – 9 |x – 2| < –2 Additional Example 4B: Special Cases of Absolute-Value Inequalities Solve the inequality. |x – 2| + 9 < 7 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.

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

26. |x| – 9 ≥ –11 +9 ≥ +9 |x| ≥ –2 Check It Out! Example 4a Solve the inequality. |x| – 9 ≥ –11 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.

27. 4|x – 3.5| ≤ –8 4 4 |x – 3.5| ≤ –2 Check It Out! Example 4b Solve the inequality. 4|x – 3.5| ≤ –8 Divide both sides by 4. 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.

28. –10 –5 5 0 10 –6 –5 –4 –3 –2 –1 0 Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. 3|x| > 15 x < –5 or x > 5 2. |x + 3| + 1 < 3 –5 < x < –1 3. A number, n, is no more than 7 units away from 5. Write and solve an inequality to show the range of possible values for n. |n– 5| ≤ 7; –2 ≤ n ≤ 12

29. Lesson Quiz: Part II Solve each inequality. ø 4. |3x| + 1 < 1 5. |x + 2| – 3 ≥ – 6 all real numbers