Solving Absolute Value Equations

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Presentation transcript:

Solving Absolute Value Equations LESSON 1–4 Solving Absolute Value Equations

A2.6(D) Formulate absolute value linear equations. Targeted TEKS A2.6(D) Formulate absolute value linear equations. A2.6(E) Solve absolute value linear equations. Mathematical Processes A2.1(B), A2.1(D) TEKS

absolute value empty set constraint extraneous solution Vocabulary

Concept

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

Solving Absolute Value Equations LESSON 1–4 Solving Absolute Value Equations