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10TH EDITION COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER 1.8 - 1 1.8 Absolute Value Equations and Inequalities Absolute Value Equations Absolute Value Inequalities Special Cases Absolute Value Models for Distance and Tolerance 1.8 - 2 Distance is 3. Distance is greater than 3. Distance is 3. Distance Distance is is greater less than 3. than 3. Distance is less than 3. –3 0 3 By definition, the equation x = 3 can be solved by finding real numbers at a distance of three units from 0. Two numbers satisfy this equation, 3 and – 3. So the solution set is 3,3. 1.8 - 3 Properties of Absolute Value 1. For b 0, a b if and only if a b or a b. 2. a b if and only if a b or a b. For any positive number b: 3. a b if and only if b a b. 4. a b if and only if a b or a b. 1.8 - 4 Example 1 SOLVING ABSOLUTE VALUE EQUATIONS Solve a. 5 3 x 12 Solution For the given expression 5 – 3x to have absolute value 12, it must represent either 12 or –12 . This requires applying Property 1, with a = 5 – 3x and b = 12. 1.8 - 5 Example 1 SOLVING ABSOLUTE VALUE EQUATIONS Solve a. 5 3 x 12 Solution 5 3 x 12 5 3x 12 or 5 3x 12 Property 1 3x 7 or 3x 17 Subtract 5. 7 x 3 or 17 x 3 Divide by –3. 1.8 - 6 Example 1 SOLVING ABSOLUTE VALUE EQUATIONS Solve a. 5 3 x 12 Solution 17 7 or x x Divide by –3. 3 3 Check the solutions by substituting them in the original absolute value equation. The solution set is 7 17 , . 3 3 1.8 - 7 Example 1 SOLVING ABSOLUTE VALUE EQUATIONS Solve b. 4 x 3 x 6 Solution 4x 3 x 6 4x 3 x 6 or 4 x 3 ( x 6) Property 2 3x 9 or 4x 3 x 6 or 5x 3 3 3 The solutionset is ,3 . x 5 5 x 3 1.8 - 8 Example 2 SOLVING ABSOLUTE VALUE INEQUALITIES Solve a. 2 x 1 7 Solution Use Property 3, replacing a with 2x + 1 and b with 7. 2x 1 7 7 2x 1 7 8 2x 6 4 x 3 Property 3 Subtract 1 from each part. Divide each part by 2. 1.8 - 9 Example 2 SOLVING ABSOLUTE VALUE INEQUALITIES Solve a. 2 x 1 7 Solution 4 x 3 Divide each part by 2. The final inequality gives the solution set (–4, 3). 1.8 - 10 Example 2 SOLVING ABSOLUTE VALUE INEQUALITIES Solve b. 2 x 1 7 Solution 2x 1 7 2x 1 7 or 2x 8 or x 4 or 2x 1 7 2x 6 x 3 Property 4 Subtract 1 from each side. Divide each part by 2. 1.8 - 11 Example 2 SOLVING ABSOLUTE VALUE INEQUALITIES Solve b. 2 x 1 7 Solution x 4 or x 3 Divide each part by 2. The solution set is ( , 4) (3, ). 1.8 - 12 Example 3 SOLVING AN ABSOLUTE VALUE INEQUALITY REQUIRING A TRANSFORMATION Solve 2 7 x 1 4. Solution 2 7x 1 4 Add 1 to each side. 2 7x 5 2 7x 5 or 7x 7 x 1 or or 2 7x 5 7x 3 3 x 7 Property 4 Subtract 2. Divide by –7; reverse the direction of each inequality. 1.8 - 13 Example 3 SOLVING AN ABSOLUTE VALUE INEQUALITY REQUIRING A TRANSFORMATION Solve 2 7 x 1 4. Solution x 1 or 3 x 7 Divide by –7; reverse the direction of each inequality. 3 The solution set is , 1, . 7 1.8 - 14 Example 4 SOLVING SPECIAL CASES OF ABSOLUTE VALUE EQUATIONS AND INEQULAITIES Solve a. 2 5 x 4 Solution Since the absolute value of a number is always nonnegative, the inequality is always true. The solution set includes all real numbers. 1.8 - 15 Example 4 SOLVING SPECIAL CASES OF ABSOLUTE VALUE EQUATIONS AND INEQULAITIES Solve b. 4 x 7 3 Solution There is no number whose absolute value is less than –3 (or less than any negative number). The solution set is . 1.8 - 16 Example 4 SOLVING SPECIAL CASES OF ABSOLUTE VALUE EQUATIONS AND INEQULAITIES Solve c. 5 x 15 0 Solution The absolute value of a number will be 0 only if that number is 0. Therefore, 5 x 15 0 is equivalent to 5x 15 0 which has solution set {–3}. Check by substituting into the original equation. 1.8 - 17 Example 5 USING ABSOLUTE INEQUALITIES TO DESCRIBE DISTANCES Write each statement using an absolute value inequality. a. k is no less than 5 units from 8. Solution Since the distance from k to 8, written k – 8 or 8 – k , is no less than 5, the distance is greater than or equal to 5. This can be written as k 8 5, or equivalently 8 k 5. 1.8 - 18 Example 5 USING ABSOLUTE INEQUALITIES TO DESCRIBE DISTANCES Write each statement using an absolute value inequality. b. n is within .001 unit of 6. Solution This statement indicates that the distance between n and 6 is less than .001, written n 6 .001 or equivalently 6 n .001. 1.8 - 19 Example 6 USING ABSOLUTE VALUE TO MODEL TOLERANCE Suppose y = 2x + 1 and we want y to be within .01 unit of 4. For what values of x will this be true? Solution y 4 .01 2 x 1 4 .01 Write an absolute value inequality. Substitute 2x + 1 for y. 2 x 3 .01 .01 2x 3 .01 Property 3 2.99 2x 3.01 Add three to each part. 1.495 x 1.505 Divide each part by 2. 1.8 - 20