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

2. 2-5 Solving Equations Involving Absolute Value

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

4. 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

5. 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| – 7 = –2
= |–2| = 2
= 17 Simplify.
Example 1

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

7. 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

8. 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 Case 2
2x – 1 = x – 1 = –7
2x – = Add 1 to each side x – = –7 + 1
2x = Simplify x = –6
Divide each side by 2.
x = Simplify x = –3
Example 2

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

10. 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

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

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

13. 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

14. 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

15. 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