Bell Ringer: Simplify each expression 1. 𝟏𝟓 2. −𝟐𝟕𝟐 3. −𝟖+𝟏𝟎 4. −𝟑−𝟗

2-5 Solving Equations Involving Absolute Value

Objectives I can solve equations using absolute value I can solve absolute value equations with coefficients represented by letters

Absolute Value The distance a number is from 0 Distance is always positive, so absolute value is always positive We can use absolute value to solve equations by plugging in a number for a given variable

Evaluate |a – 7| + 15 if a = 5. |a – 7| + 15 Expressions with Absolute Value Evaluate |a – 7| + 15 if a = 5. |a – 7| + 15 = |5 – 7| + 15 Plug in 5 for a . = |–2| + 15 5 – 7 = –2 = 2 + 15 |–2| = 2 = 17 Simplify. Example 1

Evaluate |17 – b| + 23 if b = 6. A. 17 B. 24 C. 34 D. 46 Example 1

This is the Solution Set: {-13, 13} Absolute Value Equations Case 1: The expression inside the absolute value symbol is positive or zero Case 2: The expression inside the absolute value symbol is negative Example: 𝒅 =𝟏𝟎, so d = 10 or d = -10 If 𝒙 =𝟏𝟑, what does x = ? This is the Solution Set: {-13, 13} Concept

A. Solve |2x – 1| = 7. Then graph the solution set. Solve Absolute Value Equations A. Solve |2x – 1| = 7. Then graph the solution set. |2x – 1| = 7 Original equation Case 1 Case 2 2x – 1 = 7 2x – 1 = –7 2x – 1 + 1 = 7 + 1 Add 1 to each side. 2x – 1 + 1 = –7 + 1 2x = 8 Simplify. 2x = –6 Divide each side by 2. x = 4 Simplify. x = –3 Example 2

Solve Absolute Value Equations Answer: {–3, 4} Example 2

Solve Absolute Value Equations B. Solve |p + 6| = –5. Then graph the solution set. |p + 6| = –5 means the distance between p and 6 is –5. Since distance cannot be negative, the solution is the empty set Ø. Example 2

A. Solve |2x + 3| = 5. Graph the solution set. B. {1, 4} C. {–1, –4} D. {–1, 4} Example 2

B. Solve |x – 3| = –5. A. {8, –2} B. {–8, 2} C. {8, 2} D. Example 2

Write an equation involving absolute value for the graph. Write an Absolute Value Equation Write an equation involving absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1. Example 4

Answer: |y – 1| = 5 The distance from 1 to –4 is 5 units. Write an Absolute Value Equation The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. So, an equation is |y – 1| = 5. Answer: |y – 1| = 5 Example 4

Write an equation involving the absolute value for the graph. A. |x – 2| = 4 B. |x + 2| = 4 C. |x – 4| = 2 D. |x + 4| = 2 Example 4