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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

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

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

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

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

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

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

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

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

absolute value empty set constraint extraneous solution Vocabulary

Concept

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

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

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

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

|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

A. B. C. D. Example 3

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

One Solution Check:  Answer: Example 4

A. B. C. D. Example 4

End of the Lesson