11.|–4 | |–3 – 4 | 7 ALGEBRA 1 LESSON 10-5 (For help, go to Lessons 1-3 and 1-4.) Simplify. 1.|15|2.|–3|3.|18 – 12| 4.–|–7|5.|12 – (–12)|6.|–10 + 8| Complete each statement with. 7.|3 – 7| 48.|–5| |7| – |6 – 2 | Absolute Value Equations and Inequalities

ALGEBRA 1 LESSON |15| = 152.|–3| = 3 3. |18 – 12| = |6| = 64.–|–7| = –(7) = –7 5. |12 – (–12)| = | | = |24| = 24 6.|–10 + 8| = |–2| = 2 7.|3 – 7| 4 |–4| 4 4 = 4 8.|–5| > 6 Absolute Value Equations and Inequalities Solutions 10-5

ALGEBRA 1 LESSON 10-5 Solutions 10.|6 – 2 | 3 |3 | 3 3 > |7| – – < 8 11.|–4 | > |–3 – 4 | 7 |–7 | 7 7 = Absolute Value Equations and Inequalities 10-5

ALGEBRA 1 LESSON 10-5 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 10-5

ALGEBRA 1 LESSON 10-5 Solve |3c – 6| = 9. The value of c is 5 or –1. Absolute Value Equations and Inequalities c – 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

ALGEBRA 1 LESSON 10-5 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 10-5

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. ALGEBRA 1 LESSON 10-5 greatest difference between actual and ideal Relate:0.007 mm is at most Define:Let d = actual diameter in millimeters of the piston. Write:| d – |0.007 < Absolute Value Equations and Inequalities 10-5

(continued) ALGEBRA 1 LESSON 10-5 The actual diameter must be between mm and mm, inclusive d Simplify. << Add – d – << |d – | < –0.007 d – Write a compound inequality. << Absolute Value Equations and Inequalities 10-5

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 ALGEBRA 1 LESSON 10-5 Absolute Value Equations and Inequalities 10-5