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Section 7

Chapter 2. Section 7. Absolute Value Equations and Inequalities. Use the distance definition of absolute value. Solve equations of the form | ax + b | = k , for k > 0. Solve inequalities of the form | ax + b | < k and of the form | ax + b | > k, for k > 0.

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Section 7

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  1. Chapter 2 Section 7

  2. Absolute Value Equations and Inequalities Use the distance definition of absolute value. Solve equations of the form |ax + b| = k, for k > 0. Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0. Solve absolute value equations that involve rewriting. Solve equations of the form |ax + b| = |cx + d| . Solve special cases of absolute value equations and inequalities. 2.7

  3. Use the distance definition of absolute value. Objective 1 Slide 2.7- 3

  4. The absolute valueof a number x, written |x|, is the distance from x to 0 on the number line. For example, the solutions of |x| = 5 are 5 and 5, as shown below. We need to understand the concept of absolute value in order to solve equations or inequalities involving absolute values. We solve them by solving the appropriate compound equation or inequality. Distance is 5, so |5| = 5. Distance is 5, so |5| = 5. Use the distance definition of absolute value. Slide 2.7- 4

  5. Use the distance definition of absolute value. Slide 2.7- 5

  6. Solve equations of the form |ax + b| = k, for k > 0. Objective 2 Slide 2.7- 6

  7. Remember that because absolute value refers to distance from the origin, an absolute value equation will have two parts. Use the distance definition of absolute value. Slide 2.7- 7

  8. Solve |3x – 4| = 11. 3x – 4 = 11 or 3x – 4 = 11 3x – 4 + 4 = 11 + 4 3x – 4 + 4 = 11 + 4 3x = 7 3x = 15 x = 5 Check by substituting and 5 into the original absolute value equation to verify that the solution set is CLASSROOM EXAMPLE 1 Solving an Absolute Value Equation Solution: Slide 2.7- 8

  9. Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0. Objective 3 Slide 2.7- 9

  10. Solve |3x – 4|  11. 3x – 4 ≤11 or 3x – 4  11 3x – 4 + 4 ≤ 11 + 4 3x – 4 + 4  11 + 4 3x≤ 7 3x  15 x  5 Check the solution. The solution set is The graph consists of two intervals. -5 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 8 CLASSROOM EXAMPLE 2 Solving an Absolute Value Inequality with > Solution: ] [ Slide 2.7- 10

  11. Solve |3x – 4| ≤ 11. 11 ≤ 3x – 4 ≤ 11 11 + 4 ≤ 3x – 4 ≤ 11+ 4 7 ≤ 3x≤ 15 Check the solution. The solution set is The graph consists of a single interval. -5 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 8 CLASSROOM EXAMPLE 3 Solving an Absolute Value Inequality with < Solution: [ ] Slide 2.7- 11

  12. When solving absolute value equations and inequalities of the types in Examples 1, 2, and 3, remember the following: 1. The methods describe 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. Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k, for k > 0. Slide 2.7- 12

  13. Solve absolute value equations that involve rewriting. Objective 4 Slide 2.7- 13

  14. Solve |3x + 2| + 4 = 15. First get the absolute value alone on one side of the equals sign. |3x + 2| + 4 = 15 |3x + 2| + 4 – 4 = 15 – 4 |3a + 2| = 11 3x + 2 = 11 or 3x + 2 = 11 3x = 13 3x = 9 x = 3 The solution set is CLASSROOM EXAMPLE 4 Solving an Absolute Value Equation That Requires Rewriting Solution: Slide 2.7- 14

  15. Solve the inequality. |x + 2| – 3 > 2 |x + 2| – 3 > 2 |x + 2| > 5 x + 2 > 5 or x + 2 <  5 x > 3 x < 7 Solution set: (, 7)  (3, ) CLASSROOM EXAMPLE 5 Solving Absolute Value Inequalities That Require Rewriting Solution: Slide 2.7- 15

  16. Solve the inequality. |x + 2| – 3 < 2 |x + 2| < 5 5<x + 2 < 5 7<x < 3 Solution set: (7, 3) CLASSROOM EXAMPLE 5 Solving Absolute Value Inequalities That Require Rewriting (cont’d) Solution: Slide 2.7- 16

  17. Solve equations of the form |ax + b| = |cx + d| . Objective 5 Slide 2.7- 17

  18. Solve equations of the form |ax + b| = |cx + d|. 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). Slide 2.7- 18

  19. Solve |4x – 1| = |3x + 5|. 4x – 1 = 3x + 5 or 4x – 1 = (3x + 5) 4x – 6 = 3x or 4x – 1 = 3x – 5  6 =  x or 7x =  4 x = 6 or Check that the solution set is CLASSROOM EXAMPLE 6 Solving an Equation with Two Absolute Values Solution: Slide 2.7- 19

  20. Solve special cases of absolute value equations and inequalities. Objective 6 Slide 2.7- 20

  21. Solve special cases of absolute value equations and inequalities. Special Cases of Absolute Value 1. The absolute value of an expression can never be negative; that is, |a|  0 for all real numbers a. 2. The absolute value of an expression equals 0 only when the expression is equal to 0. Slide 2.7- 21

  22. Solve each equation. |6x + 7| = –5 The absolute value of an expression can never be negative, so there are no solutions for this equation. The solution set is . The expression will equal 0 only if The solution of the equation is 12. The solution set is {12}, with just one element. CLASSROOM EXAMPLE 7 Solving Special Cases of Absolute Value Equations Solution: Slide 2.7- 22

  23. Solve each inequality. |x | > – 1 The absolute value of a number is always greater than or equal to 0. The solution set is (, ). |x – 10| – 2 ≤ –3 |x – 10| ≤ –1 Add 2 to each side. There is no number whose absolute value is less than –1, so the inequality has no solution.The solution set is . |x + 2| ≤ 0 The value of |x + 2| will never be less than 0. |x + 2| will equal 0 when x = –2. The solution set is {–2}. CLASSROOM EXAMPLE 8 Solving Special Cases of Absolute Value Inequalities Solution: Slide 2.7- 23

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