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Lesson Menu 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

Over Lesson 1–3 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?

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

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

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

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

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

CCSS 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.

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

Vocabulary absolute value empty set constraint extraneous solution

Concept

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.

Example 1 A.18.3 B.1.7 C.–1.7 D.–13.7

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

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

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.

A. B. C. D. Example 3

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

Example 4 One Solution Check: Answer:

A. B. C. D. Example 4

End of the Lesson