ALGEBRA 1 Lesson 3-6 Warm-Up. ALGEBRA 1 “Absolute Value Equations and Inequalities (3-6) (3-1) How do you solve absolute value equations and inequalities?

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ALGEBRA 1 Lesson 3-6 Warm-Up

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:

ALGEBRA 1 Solve and check |a| – 3 = 5. |a| – = 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 Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

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

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 – –2 + 5 > y – < Simplify.y 3 > y 7 < and Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

ALGEBRA 1 The ideal diameter of a piston for one type of car is mm. The actual diameter can vary from the ideal diameter by at most 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 – |0.007 < Absolute Value Equations and Inequalities LESSON 3-6 Additional Examples

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

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