Chapter 2 Section 7
2.7 Absolute Value Equations and Inequalities Objectives 1 2 3 4 5 6 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.
Objective 1 Use the distance definition of absolute value. Slide 2.7-3
Use the distance definition of absolute value. The absolute value of 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. Slide 2.7-4
Use the distance definition of absolute value. Slide 2.7-5
Objective 2 Solve equations of the form ax + b = k, for k > 0. Slide 2.7-6
Use the distance definition of absolute value. Remember that because absolute value refers to distance from the origin, an absolute value equation will have two parts. Slide 2.7-7
CLASSROOM EXAMPLE 1 Solve 3x 4 = 11. Solution: Solving an Absolute Value Equation 3x 4 = 11 or 3x 4 = 11 3x 4 + 4 = 11 + 4 3x 4 + 4 = 11 + 4 3x = 7 3x = 15 x = x = 5 Check by substituting and 5 into the original absolute value equation to verify that the solution set is 7 3 7 3 7,5. 3 Slide 2.7-8
Objective 3 Solve inequalities of the form ax + b < k and of the form ax + b > k, for k > 0. Slide 2.7-9
CLASSROOM EXAMPLE 2 Solve 3x 4 11. Solution: Solving an Absolute Value Inequality with > 3x 4 11 or 3x 4 11 3x 4 + 4 11 + 4 3x 4 + 4 11 + 4 3x 7 3x 15 x Check the solution. The solution set is The graph consists of two intervals. 7 3 x 5 7, [ 5, ). 3 U ] -5-4 -3-2 -1 0 1 2 3 4 [ 5 6 7 8 Slide 2.7-10
CLASSROOM EXAMPLE 3 Solve 3x 4 11. Solution: Solving an Absolute Value Inequality with < 11 3x 4 11 11 + 4 3x 4 11+ 4 7 3x 15 7 x 5 3 Check the solution. The solution set is The graph consists of a single interval. 7, 5. 3 [ -5-4 -3-2 -1 0 1 2 3 4 5 6 7 8 ] Slide 2.7-11
Solve inequalities of the form ax + b < k and of the form ax + b > k, for k > 0. 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. Slide 2.7-12
Objective 4 Solve absolute value equations that involve rewriting. Slide 2.7-13
CLASSROOM EXAMPLE 4 Solve 3x + 2 + 4 = 15. Solution: Solving an Absolute Value Equation That Requires Rewriting 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 = 13 3 x = 3 The solution set is 13, 3. 3 Slide 2.7-14
CLASSROOM EXAMPLE 5 Solve the inequality. Solving Absolute Value Inequalities That Require Rewriting x + 2 3 > 2 Solution: x + 2 3 > 2 x + 2 > 5 x + 2 > 5 or x + 2 < 5 x > 3 x < 7 Solution set: (, 7) (3, ) Slide 2.7-15
CLASSROOM EXAMPLE 5 Solve the inequality. Solving Absolute Value Inequalities That Require Rewriting (cont d) x + 2 3 < 2 Solution: x + 2 < 5 5 < x + 2 < 5 7 < x < 3 Solution set: ( 7, 3) Slide 2.7-16
Objective 5 Solve equations of the form ax + b = cx + d. Slide 2.7-17
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
CLASSROOM EXAMPLE 6 Solve 4x 1 = 3x + 5. Solution: Solving an Equation with Two Absolute Values 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 x = 4 7 Check that the solution set is 4,6. 7 Slide 2.7-19
Objective 6 Solve special cases of absolute value equations and inequalities. Slide 2.7-20
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
CLASSROOM EXAMPLE 7 Solve each equation. 6x + 7 = 5 Solving Special Cases of Absolute Value Equations Solution: The absolute value of an expression can never be negative, so there are no solutions for this equation. The solution set is. 1 3 0 4 x = 1 3 4 x 1 3 4 x = x = 12. The expression will equal 0 only if The solution of the equation is 12. The solution set is {12}, with just one element. Slide 2.7-22
CLASSROOM EXAMPLE 8 Solve each inequality. x > 1 Solution: 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 Solving Special Cases of Absolute Value Inequalities The value of x + 2 will never be less than 0. x + 2 will equal 0 when x = 2. The solution set is { 2}. Slide 2.7-23