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Absolute Value Equations and Inequalities

Absolute Value Equations and Inequalities. 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 <, =, or >. 7. |3 – 7| 4 8. |–5| + 2 6

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Absolute Value Equations and Inequalities

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  1. Absolute Value Equations and Inequalities 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 <, =, or >. 7. |3 – 7| 4 8. |–5| + 2 6 9. |7| – 1 8 10. |6 – 2 | 3 1 4 5 8 11. |–4 | + 2 2 12. |–3 – 4 | 7 2 3 1 3 1 2 1 8 1 2 5 8 10-5

  2. Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 Solutions 1. |15| = 15 2. |–3| = 3 3. |18 – 12| = |6| = 6 4. –|–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 10-5

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

  4. Check: |a| – 3 = 5 |8| – 3 5 Substitute 8 and –8 for a. |–8| – 3 5 8 – 3 = 5 8 – 3 = 5 Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 Solve and check |a| – 3 = 5. |a| – 3 + 3 = 5 + 3Add 3 to each side. |a| = 8 Simplify. a = 8 or a = –8 Definition of absolute value. 10-5

  5. 3c – 6 = 9 Write two equations. 3c – 6 = –9 3c – 6 + 6 = 9 + 6Add 6 to each side. 3c – 6 + 6 = –9 + 6 3c = 15 3c = –3 Divide each side by 3. 3c 3 15 3 3c 3 –3 3 c = 5 c = –1 = = Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 Solve |3c – 6| = 9. The value of c is 5 or –1. 10-5

  6. < < < < < < and Write a compound inequality. y – 5 –2 y – 5 2 > > > y – 5 + 5 2 + 5 y – 5 + 5 –2 + 5 Add 5 to each side. y 7 y 3 and Simplify. 3 y 7 Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 Solve |y – 5| 2. Graph the solutions. 10-5

  7. < greatest difference between actual and ideal is at most Relate: 0.007 mm Define: Let d = actual diameter in millimeters of the piston. Write: | d – 88.000| 0.007 Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 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. 10-5

  8. < < < < < < < |d – 88.000| 0.007 –0.007 d – 88.000 0.007   Write a compound inequality. Add 88.000. –0.007 + 88.000d – 88.000 + 88.000 0.007 + 88.000 87.993 d 88.007   Simplify. Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 (continued) The actual diameter must be between 87.993 mm and 88.007 mm, inclusive. 10-5

  9. < Solve. 1. |a| + 6 = 9 2. |2x + 3| = 7 3. |p + 6| 1 4. 3|x + 4| > 15 < < –7 p –5 Absolute Value Equations and Inequalities ALGEBRA 1 LESSON 10-5 a = 3 or a = –3 x = 2 or x = –5 x > 1 or x < –9 10-5

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