1. 11.|–4 | + 2 212.|–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| + 2 6 9.|7| – 1 810.|6 – 2 | 3 1414 2323 1313 1212 1818 1212 Absolute Value Equations and Inequalities 10-5 5858 5858

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

3. ALGEBRA 1 LESSON 10-5 Solutions 10.|6 – 2 | 3 |3 | 3 3 > 3 1414 3434 5858 5858 3434 5858 9.|7| – 1 8 7 – 1 8 6 < 8 11.|–4 | + 2 2 4 + 2 2 7 > 2 2323 1313 1212 2323 1313 1212 1212 12.|–3 – 4 | 7 |–7 | 7 7 = 7 1818 1212 5858 5858 5858 5858 5858 Absolute Value Equations and Inequalities 10-5

4. ALGEBRA 1 LESSON 10-5 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 10-5

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

6. 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 – 5 + 5 –2 + 5 > y – 5 + 5 2 + 5 < Simplify.y 3 > y 7 < and Absolute Value Equations and Inequalities 10-5

7. 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. 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 – 88.000|0.007 < Absolute Value Equations and Inequalities 10-5

8. (continued) ALGEBRA 1 LESSON 10-5 The actual diameter must be between 87.993 mm and 88.007 mm, inclusive. 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 10-5

9. 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