1. Algebra 2 Lesson 1-5 (Page 33) ALGEBRA 2 LESSON 1-5 Absolute Value Equations and Inequalities 1-1

2. To solve absolute value equations. ALGEBRA 2 LESSON 1-5 Absolute Value Equations and Inequalities 1-1

3. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve |15 – 3x| = 6. |15 – 3x|= 6 15 – 3x= 6or15 – 3x = –6The value of 15 – 3x can be 6 or – 6 since |6| and |–6| both equal 6. x= 3or x = 7Divide each side of both equations by –3. –3x= –9 –3x = –21Subtract 15 from each side of both equations. Check:|15 – 3x|= 6|15 – 3x|=6 |15 – 3(3)|6 |15 – 3(7)|6 |6|6 |–6|6 6 = 6 6 =6 1-5

4. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Check your answer. 1-5

5. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve 4 – 2|x + 9| = –5. 4 – 2|x + 9| = –5 –2|x + 9| = –9Add –4 to each side. x = –4.5 or x = –13.5Subtract 9 from each side of both equations. Check:4 – 2 |x + 9|=–54 – 2|x + 9| = –5 4 – 2 |–4.5 + 9|–54 – 2 |–13.5 + 9| –5 4 – 2 |4.5|–54 – 2 |–4.5|–5 4 – 2 (4.5)–54 – 2 (4.5)–5 –5 =–5–5 = –5 1-5 |x + 9| = Divide each side by –2. 9292 x + 9 = or x + 9 = –Rewrite as two equations. 9292 9292

6. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Check your answer. 1-5

7. New Vocabulary ALGEBRA 2 LESSON 1-5 Absolute Value Equations and Inequalities 1-1 An extraneous solution is a solution of an equation derived from an original equation that is not a solution of the original equation.

8. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve |3x – 4| = –4x – 1. |3x – 4| = –4x – 1 1-5 3x – 4 =–4x – 1or3x – 4=–(–4x –1) Rewrite as two equations. 7x – 4 =– 13x – 4=4x + 1 Solve each equation. 7x =3–x=57x =3–x=5 x =orx=–5 3737

9. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 (continued) Check:|3x – 4|= –4x – 1|3x – 4| = –4x – 1 |3 ( ) – 4| –4 ( ) – 1|3(–5) – 4| (–4(–5) –1) 3737 3737 |– |–|–19|19 19 7 19 7 –19=19 19 7 19 7 1-5 =/ The only solution is –5. is an extraneous solution. 3737

10. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Check for extraneous solutions. 1-5

11. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Check for extraneous solutions. 1-5

12. To solve absolute value inequalities. ALGEBRA 2 LESSON 1-5 Absolute Value Equations and Inequalities 1-1

13. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve |2x – 5| > 3. Graph the solution. |2x – 5| > 3 2x – 5 3Rewrite as a compound inequality. x 4 2x 8Solve for x. 1-5

14. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Graph the solution. 1-5

15. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve –2|x + 1| + 5 –3. Graph the solution. 1-5 > – |x + 1| 4Divide each side by –2 and reverse the inequality. < – –2|x + 1| + 5 –3 > – –2|x + 1| –8Isolate the absolute value expression. Subtract 5 from each side. > – –4x + 1 4Rewrite as a compound inequality. < – < – –5x 3Solve for x. < – < –

16. Absolute Value Equations and Inequalities ALGEBRA 2 LESSON 1-5 Solve. Graph the solution. 1-5

17. New Vocabulary ALGEBRA 2 LESSON 1-5 Absolute Value Equations and Inequalities 1-1 The tolerance equals one-half of the difference between the maximum and the minimum values.

18. Absolute Value Equations and Inequalities The area A in square inches of a square photo is required to satisfy 8.5 A 8.9. Write this requirement as an absolute value inequality. ALGEBRA 2 LESSON 1-5 1-5 < – < – Write an inequality.–0.2 A – 8.7 0.2 < – < – Rewrite as an absolute value inequality.|A – 8.7| 0.2 < – Find the tolerance. 8.9 – 8.5 2 = 0.4 2 = 0.2Find the average of the maximum and minimum values. 8.9 + 8.5 2 = 17.4 2 = 8.7

19. Absolute Value Equations and Inequalities The specification for the circumference C in inches of a Basketball for junior high school is 27.75 C 28.5. Write the Specification as an absolute value inequality. ALGEBRA 2 LESSON 1-5 1-5 < – < –