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Learn to evaluate expressions with absolute values and solve absolute value equations. Practice solving verbal and real-world problems.

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  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? A. 23 + 2(65) = 65 B. 23 + n = 65 C. 23 = 2n + 65 D. 23 + 2n = 65 5-Minute Check 3

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

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

  7. 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. 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. ? ? |5 + 3| = 8 |–11 + 3| = 8 ? ? |8| = 8 |–8| = 8 8 = 8 8 = 8   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 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? A. {5} B. {–10, 5} C. {–5, 10} D. {–5} Example 2

  15. 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

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