# 9.2 Absolute Value Equations and Inequalities

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9.2 Absolute Value Equations and Inequalities

Use the distance definition of absolute value.
Objective 1 Use the distance definition of absolute value. Slide

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

Use the distance definition of absolute value.
Slide

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

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

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

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

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

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

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

Solve absolute value equations that involve rewriting.
Objective 4 Solve absolute value equations that involve rewriting. Slide

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

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

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

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

Solving |ax + b| = |cx + d|
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

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

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

Special Cases of Absolute Value
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

Solve each equation. |6x + 7| = – 5
CLASSROOM EXAMPLE 7 Solving Special Cases of Absolute Value Equations 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 . Solution: The expression will equal 0 only if The solution of the equation is 12. The solution set is {12}, with just one element. Slide

Solve each inequality. |x | > – 1
CLASSROOM EXAMPLE 8 Solving Special Cases of Absolute Value Inequalities 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}. Solution: Slide