1. Splash Screen

2. Five-Minute Check (over Lesson 1–3) CCSS Then/Now New Vocabulary
Key Concept: Absolute Value
Example 1: Evaluate an Expression with Absolute Value
Example 2: Real-World Example: Solve an Absolute Value Equation
Example 3: No Solution
Example 4: One Solution
Lesson Menu

3. Which algebraic expression represents the verbal expression five less than the product of the cube of a number and –4?
A. 5 – (–4n3)
B. –4n3 – 5
C. –4n3 + 5
D. n3 – 5
5-Minute Check 2

4. Which equation represents the verbal expression the sum of 23 and twice a number is 65?
B n = 65
C. 23 = 2n + 65
D n = 65
5-Minute Check 3

5. Solve the equation 12f – 4 = 7 + f.
B. 0.5
C. 0
D. –1
5-Minute Check 4

6. Solve the equation 10y + 1 = 3(–2y – 5).
B. 1
C. 0
D. –1
5-Minute Check 5

7. Mathematical Practices 6 Attend to precision.
Content Standards
A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity.
A.CED.1 Create equations and inequalities in one variable and use them to solve problems.
Mathematical Practices
6 Attend to precision.
CCSS

8. You solved equations using properties of equality.
Evaluate expressions involving absolute values.
Solve absolute value equations.
Then/Now

9. absolute value
empty set
constraint
extraneous solution
Vocabulary

10. Concept

11. Replace x with 4. Multiply 2 and 4 first. Subtract 8 from 6. Add.
Evaluate an Expression with Absolute Value
Replace x with 4.
Multiply 2 and 4 first.
Subtract 8 from 6.
Add.
Answer: 4.7
Example 1

12. A. 18.3
B. 1.7
C. –1.7
D. –13.7
Example 1

13. Case 1 a = b y + 3 = 8 y + 3 – 3 = 8 – 3 y = 5 Case 2 a = –b
Solve an Absolute Value Equation
Case 1 a = b
y + 3 = 8
y + 3 – 3 = 8 – 3
y = 5
Case 2 a = –b
y + 3 = –8
y + 3 – 3 = –8 – 3
y = –11
Check |y + 3| = 8
|y + 3| = 8 ?
|5 + 3| = 8 ?
|–11 + 3| = 8 ?
|8| = 8 ?
|–8| = 8 
8 = 8 
8 = 8
Answer: The solutions are 5 and –11. Thus, the solution set is –11, 5.
Example 2

14. What is the solution to |2x + 5| = 15?
B. {–10, 5}
C. {–5, 10}
D. {–5}
Example 2

15. |6 – 4t| + 5 = 0 Original equation
No Solution
Solve |6 – 4t| + 5 = 0.
|6 – 4t| + 5 = 0 Original equation
|6 – 4t| = –5 Subtract 5 from each side.
This sentence is never true.
Answer: The solution set is .
Example 3

16. A. B. C. D.
Example 3

17. One Solution
Case 1 a = b
8 + y = 2y – 3
8 = y – 3
11 = y
Example 4

18. One Solution
Check: 
Answer:
Example 4

19. A. B. C. D.
Example 4

20. End of the Lesson

21. Pages 30 – 31
# 15 – 25 ODD, 31, 32, 37 – 41 ODD, 45