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Presentation on theme: "Splash Screen."— Presentation transcript:

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 three times the sum of a number and its square? A. 3(x2) B. 3x + x2 C. 3(x + x2) D. 3 + x + x2 5-Minute Check 1

4 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

5 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

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

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

8 A. B. C. D. 5-Minute Check 6

9 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

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

11 absolute value empty set constraint extraneous solution Vocabulary

12 Concept

13 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

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

15 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

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

17 |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

18 A. B. C. D. Example 3

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

20 One Solution Check: Answer: Example 4

21 A. B. C. D. Example 4

22 End of the Lesson


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