1. Splash Screen

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

3. Over Lesson 1–3 A.A B.B C.C D.D 5-Minute Check 1 A.3(x 2 ) B.3x + x 2 C.3(x + x 2 ) D.3 + x + x 2 Which algebraic expression represents the verbal expression three times the sum of a number and its square?

4. Over Lesson 1–3 A.A B.B C.C D.D 5-Minute Check 2 A.5 – (–4n 3 ) B.–4n 3 – 5 C.–4n 3 + 5 D.n 3 – 5 Which algebraic expression represents the verbal expression five less than the product of the cube of a number and –4?

5. Over Lesson 1–3 A.A B.B C.C D.D 5-Minute Check 3 A.23 + 2(65) = 65 B.23 + n = 65 C.23 = 2n + 65 D.2n + 23 = 65 Which equation represents the verbal expression the sum of 23 and twice a number is 65?

6. Over Lesson 1–3 A.A B.B C.C D.D 5-Minute Check 4 A.1 B.0.5 C.0 D.–1 Solve the equation 12f – 4 = 7 + f.

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

8. Over Lesson 1–3 A.B. C.D. A.A B.B C.C D.D 5-Minute Check 6

9. Then/Now You solved equations using properties of equality. (Lesson 1–3) Evaluate expressions involving absolute values. Solve absolute value equations.

10. Vocabulary absolute value empty set extraneous solution

11. Concept

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

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

14. Example 2 Solve an Absolute Value Equation Case 1a=b y + 3=8 y + 3 – 3=8 – 3 y=5y=5 Answer: The solutions are 5 and –11. Thus, the solution set is  –11, 5 . Check|y + 3|=8 Case 2a=–b y + 3=–8 y + 3 – 3=–8 – 3 y=–11 |y + 3|=8 ? |5 + 3|=8 ? |8|=8 8=88=8 ? |–11 + 3|=8 ? |–8|=8 8=88=8

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

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

17. A. B. C. D. A.A B.B C.C D.D Example 3

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

19. Example 4 One Solution Check: Answer:

20. A. B. C. D. A.A B.B C.C D.D Example 4

21. End of the Lesson