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Learn how to evaluate expressions with absolute value and solve absolute value equations. Discover the concept of absolute value and distinguish between no solution and one solution. Includes multiple examples and a lesson menu.
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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 Lesson Menu
A B C D 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
A B C D 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
A B C D 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. 2n + 23 = 65 5-Minute Check 3
A B C D Solve the equation 12f – 4 = 7 + f. A. 1 B. 0.5 C. 0 D. –1 5-Minute Check 4
A B C D Solve the equation 10y + 1 = 3(–2y – 5). A. 2 B. 1 C. 0 D. –1 5-Minute Check 5
A B C D A.B. C.D. 5-Minute Check 6
You solved equations using properties of equality. (Lesson 1–3) • Evaluate expressions involving absolute values. • Solve absolute value equations. Then/Now
absolute value • empty set • extraneous solution Vocabulary
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 B C D A. 18.3 B. 1.7 C. –1.7 D. –13.7 Example 1
? ? |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
A B C D What is the solution to |2x + 5| = 15? A. {5} B. {–10, 5} C. {–5, 10} D. {–5} Example 2
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 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 A. B. C. D. Example 4