Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations and Inequalities Involving Absolute Value Learn to solve equations involving absolute value. Learn to solve inequalities involving absolute value. SECTION

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley ABSOLUTE VALUE

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley THE SOLUTIONS OF

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Solving an Equation Involving Absolute Value Solve each equation. The solution set is {–3}. Solution Check the solution.  ?

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Solving an Equation Involving Absolute Value The solution set is {–5, 8}. Solution continued We leave it to you to check the solutions. Isolate the absolute value.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Solving an Equation of the Form If |u| = |v|, then either u is equal to |v| or u is equal to –|v|. Since |v| = ± v in every case, we have u = v or u = –v. Thus, Solution Solve The solution set is {–2}. We leave the check to you.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley RULES FOR SOLVING ABSOLUTE VALUE INEQUALITIES If a > 0, and u is an algebraic expression, then

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solving an Inequality Involving an Absolute Value Solve the inequalityand graph the the solution set. Rule 2 applies here, with u = 4x – 1 and a = 9. Solution

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solving an Inequality Involving an Absolute Value Solution continued The solution set isthat is, the solution set is the closed interval 120–1–2 ] [ 3–3

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 In the introduction to this section, we wanted to find the possible search range (in miles) for a search plane that has 30 gallons of fuel and uses 10 gallons of fuel per hour. We were told that the search plane normally averages 110 miles per hour, but that weather conditions could affect the average speed by as much as 15 miles per hour (either slower or faster). How do we find the possible search range? Finding the Search Range of an Aircraft

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 To find distance, we need both speed and time. Let x = actual speed in mph We know actual speed is within 15 mph of average speed, 110 mph. That is, |Actual speed – Average speed| ≤ 15 mph Finding the Search Range of an Aircraft Solution The actual speed of the search plane is between 95 and 125 mph.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 The plane uses 10 g of fuel per hour. It has 30 g, so it can fly for 3 hr. So the actual number of miles the search plane can fly is 3x. Finding the Search Range of an Aircraft Solution continued The search plane’s range is between 285 and 375 miles.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Solving an Inequality Involving an Absolute Value Solve the inequalityand graph the the solution set. Rule 4 applies here, with u = 2x – 8 and a = 9. Solution

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Solving an Inequality Involving an Absolute Value The solution set is {x | x ≤ 2 or x ≥ 6}; Solution continued x ≤ 2 or x ≥ 6 (–∞, 2] U [6, ∞) [] in interval notation it is (–∞, 2] U [6, ∞).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Solving Special Cases of Absolute Value Inequalities Solve each inequality. a. The absolute value is always nonnegative, so |3x – 2| > –5 is true for all real numbers x. The solution set is all real numbers, or (–∞, ∞). Solution b. There is no real number with absolute value ≤ –2. The solution set is the empty set, or