1. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 2 Linear Inequalities and Absolute Value Chapter 3

3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 3.3 - 3 3.3 Absolute Value Equations and Inequalities

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 4 3.3 Absolute Value Equations and Inequalities Objectives 1. Use the distance definition of absolute value. 2. Solve equations of the form | ax + b | = k, for k > 0. 3. Solve inequalities of the form | ax + b | k, for k > 0. 4. Solve absolute value equations that involve rewriting. 5. Solve equations of the form | ax + b | = | cx + d |. 6. Solve special cases of absolute value equations and inequalities.

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 5 3.3 Absolute Value Equations and Inequalities Absolute Value In Section1.1 we saw that the absolute value of a number x, written | x |, represents the distance from x to 0 on the number line. 05–5 For example, the solutions of | x | = 5 are 5 and –5. 5 units from zero. x = –5 or x = 5

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 6 3.3 Absolute Value Equations and Inequalities Absolute Value 05–5 Because the absolute value represents the distance from 0, it is reasonable to interpret the solutions of | x | > 5 to be all numbers that are more than 5 units from 0. The set (- ∞, –5) U (5, ∞ ) fits this description. Because the graph consists of two separate intervals, the solution set is described using or as x < –5 or x > 5. More than 5 units from zero More than 5 units from zero

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 7 3.3 Absolute Value Equations and Inequalities Absolute Value 05–5 The solution set of | x | < 5 consists of all numbers that are less than 5 units from 0 on the number line. Another way of thinking of this is to think of all numbers between –5 and 5. This set of numbers is given by (–5, 5), as shown in the figure below. Here, the graph shows that –5 < x < 5, which means x > –5 and x < 5. Less than 5 units from zero

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 8 3.3 Absolute Value Equations and Inequalities Absolute Value The equation and inequalities just described are examples of absolute value equations and inequalities. They involve the absolute value of a variable expression and generally take the form |ax + b| = k,|ax + b| > k,|ax + b| < k, or where k is a positive number. | x | = 5 has the same solution set as x = –5 or x = 5, | x | > 5 has the same solution set as x 5, | x | –5 and x < 5. From the previous examples, we see that

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 9 3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities 1. To solve | ax + b | = k, solve the following compound equation. Let k be a positive real number, and p and q be real numbers. ax + b = k or ax + b = – k. The solution set is usually of the form {p, q}, which includes two numbers. pq

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 10 3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities 2. To solve | ax + b | > k, solve the following compound inequality. Let k be a positive real number, and p and q be real numbers. ax + b > k or ax + b < – k. The solution set is of the form (- ∞, p ) U ( q, ∞ ), which consists of two separate intervals. pq

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 11 3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities 3. To solve | ax + b | < k, solve the three-part inequality Let k be a positive real number, and p and q be real numbers. – k < ax + b < k The solution set is of the form ( p, q ), a single interval. pq

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 12 3.3 Absolute Value Equations and Inequalities EXAMPLE 1Solving an Absolute Value Equation Solve |2 x + 3| = 5. For |2 x + 3| to equal 5, 2 x + 3 must be 5 units from 0 on the number line. This can happen only when 2 x + 3 = 5 or 2 x + 3 = –5. Solve this compound equation as follows. 2 x + 3 = 5 or2 x + 3 = –5 2 x = 2 x = 1 2 x = –8 x = –4 or Check by substituting 1 and then –4 in the original absolute value equation to verify that the solution set is {–4, 1}. –5–4–3–2–1012345

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 13 3.3 Absolute Value Equations and Inequalities Writing Compound Statements NOTE Some people prefer to write the compound statements in Cases 1 and 2 of the summary on the previous slides as the equivalent forms ax + b = k or –(ax + b) = k and ax + b > k or –(ax + b) > k. These forms produce the same results.

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 14 3.3 Absolute Value Equations and Inequalities EXAMPLE 2Solving an Absolute Value Inequality with > Solve |2 x + 3| > 5. By part 2 of the summary, this absolute value inequality is rewritten as 2 x + 3 > 5 or2 x + 3 < –5 2 x > 2 x > 1 2 x < –8 x < –4 or because 2 x + 3 must represent a number that is more than 5 units from 0 on either side of the number line. Now, solve the compound inequality. –5–4–3–2–1012345 2 x + 3 > 5 or2 x + 3 < –5, Check these solutions. The solution set is (–∞, –4) U (1, ∞ ). Notice that the graph consists of two intervals.

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 15 3.3 Absolute Value Equations and Inequalities EXAMPLE 3Solving an Absolute Value Inequality with < Solve |2 x + 3| < 5. –5 < 2 x + 3 < 5. –8 < 2 x < 2 –4 < x < 1 The expression 2 x + 3 must represent a number that is less than 5 units from 0 on either side of the number line. 2 x + 3 must be between –5 and 5. As part 3 of the summary shows, this is written as the three-part inequality –5–4–3–2–1012345 Check that the solution set is (–4, 1), so the graph consists of the single interval shown below. Subtract 3 from each part. Divide each part by 2.

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 16 CAUTION When solving absolute value equations and inequalities of the types in Examples 1, 2, and 3, remember the following. 1.The methods described apply when the constant is alone on one side of the equation or inequality and is positive. 2.Absolute value equations and absolute value inequalities of the form | ax + b | > k translate into “or” compound statements. 3.Absolute value inequalities of the form | ax + b | < k translate into “and” compound statements, which may be written as three-part inequalities. 4.An “or” statement cannot be written in three parts. It would be incorrect to use –5 > 2 x + 3 > 5 in Example 2, because this would imply that –5 > 5, which is false. 3.3 Absolute Value Equations and Inequalities Caution

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 17 3.3 Absolute Value Equations and Inequalities EXAMPLE 4Solving an Absolute Value Equation That Requires Rewriting Solve the equation | x – 7| + 6 = 9. | x – 7| + 6 – 6 = 9 – 6 | x – 7| = 3 x – 7 = 3or x – 7 = –3 First, rewrite so that the absolute value expression is alone on one side of the equals sign by subtracting 6 from each side. Now use the method shown in Example 1. Subtract 6. x = 10 Check that the solution set is {4, 10} by substituting 4 and then 10 into the original equation. or x = 4

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 18 3.3 Absolute Value Equations and Inequalities Solving |ax + b| = |cx + d| To solve an absolute value equation of the form |ax + b| = |cx + d|, solve the compound equation ax + b = cx + d or ax + b = – ( cx + d ).

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 19 3.3 Absolute Value Equations and Inequalities EXAMPLE 5Solving an Equation with Two Absolute Values Solve the equation | y + 4| = |2 y – 7|. y + 4 = 2 y – 7 or y + 4 = –(2 y – 7). y + 4 = 2 y – 7 or y + 4 = –(2 y – 7) This equation is satisfied either if y + 4 and 2 y – 7 are equal to each other, or if y + 4 and 2 y – 7 are negatives of each other. Thus, Solve each equation. 11 = y Check that the solution set is {1, 11}. 3 y = 3 y = 1

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 20 3.3 Absolute Value Equations and Inequalities Special Cases for Absolute Value 1. The absolute value of an expression can never be negative: | a | ≥ 0 for all real numbers a. 2. The absolute value of an expression equals 0 only when the expression is equal to 0.

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 21 3.3 Absolute Value Equations and Inequalities EXAMPLE 6Solving Special Cases of Absolute Value Equations Solve each equation. See Case 1 in the preceding slide. Since the absolute value of an expression can never be negative, there are no solutions for this equation. The solution set is Ø. (a)|2 n + 3| = –7 See Case 2 in the preceding slide. The absolute value of the expres- sion 6 w – 1 will equal 0 only if 6 w – 1 = 0. (b)|6 w – 1| = 0 The solution of this equation is. Thus, the solution set of the original equation is { }, with just one element. Check by substitution. 1 6 1 6

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 22 3.3 Absolute Value Equations and Inequalities EXAMPLE 7Solving Special Cases of Absolute Value Inequalities Solve each inequality. The absolute value of a number is always greater than or equal to 0. Thus, | x | ≥ –2 is true for all real numbers. The solution set is (–∞, ∞ ). (a)| x | ≥ –2 Add 1 to each side to get the absolute value expression alone on one side. | x + 5| < –7 (b)| x + 5| – 1 < –8 There is no number whose absolute value is less than –7, so this inequality has no solution. The solution set is Ø.

23. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 3.3 - 23 3.3 Absolute Value Equations and Inequalities EXAMPLE 7Solving Special Cases of Absolute Value Inequalities Solve each inequality. Subtracting 2 from each side gives | x – 9| ≤ 0 (c)| x – 9| + 2 ≤ 2 The value of | x – 9| will never be less than 0. However, | x – 9| will equal 0 when x = 9. Therefore, the solution set is {9}.