120 likes | 130 Views
LESSON 1 –4. Solving Absolute Value 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.
E N D
LESSON 1–4 Solving Absolute Value 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
? ? |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
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. 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
LESSON 1–4 Solving Absolute Value Equations