1. ALGEBRA 1 Lesson 3-6 Warm-Up

2. ALGEBRA 1 “Absolute Value Equations and Inequalities (3-6) (3-1) How do you solve absolute value equations and inequalities? Note: To get rid of the absolute value “brackets” around a variable in the solution, make the solution positive and negative, because it could be either one (since the absolute value brackets make it positive regardless of the sign). Example:

3. ALGEBRA 1 Solve and check |a| – 3 = 5. |a| – 3 + 3 = 5 + 3Add 3 to each side. |a| = 8Simplify. a = 8 or a = –8Definition of absolute value. Check: |a| – 3 = 5 |8| – 3 5Substitute 8 and –8 for a.|–8| – 3 5 8 – 3 = 5 8 – 3 = 5 Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

4. ALGEBRA 1 Solve |3c – 6| = 9. The value of c is 5 or –1. 3c – 6 = 9Write two equations. 3c – 6 = –9 3c – 6 + 6 = 9 + 6Add 6 to each side.3c – 6 + 6 = –9 + 6 3c = 153c = –3 Divide each side by 3. 3c33c3 = 15 3 3c33c3 = –3 3 c = 5c = –1 Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

5. ALGEBRA 1 Solve |y – 5| 2. Graph the solutions. < 3 y 7 << Write a compound inequality.y – 5 –2 > y – 5 2 < and Add 5 to each side.y – 5 + 5 –2 + 5 > y – 5 + 5 2 + 5 < Simplify.y 3 > y 7 < and Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

6. ALGEBRA 1 The ideal diameter of a piston for one type of car is 88.000 mm. The actual diameter can vary from the ideal diameter by at most 0.007 mm. Find the range of acceptable diameters for the piston. greatest difference between actual and ideal Words:0.007 mm is at most Define:Let d = actual diameter in millimeters of the piston. Equation:| d – 88.000|0.007 < Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

7. ALGEBRA 1 (continued) The actual diameter must be between 87.993 mm and 88.007 mm, Inclusive (including the two end values). 87.993d88.007 Simplify. << Add 88.000.–0.007 + 88.000 d – 88.000 + 88.000 0.007 + 88.000 << |d – 88.000| 0.007 < –0.007 d – 88.000 0.007 Write a compound inequality. << Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

8. ALGEBRA 1 Solve. 1.|a| + 6 = 92.|2x + 3| = 7 3.|p + 6| 14.3|x + 4| > 15 < a = 3 or a = –3x = 2 or x = –5 –7 p –5 << x > 1 or x < –9 Absolute Value Equations and Inequalities LESSON 3-6 Lesson Quiz