1 / 11

1-5 Absolute Value Equations and Inequalities

1-5 Absolute Value Equations and Inequalities. Algebraic Definition of Absolute Value - If x > 0, then | x | = x - If x < 0. then | x | = -x. –3 x = –9 –3 x = –21 Subtract 15 from each side of both equations. Check: |15 – 3 x | = 6 |15 – 3 x | = 6

Download Presentation

1-5 Absolute Value Equations and Inequalities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 1-5 Absolute Value Equations and Inequalities

  2. Algebraic Definition of Absolute Value - If x > 0, then | x | = x - If x < 0. then | x | = -x

  3. –3x = –9 –3x = –21 Subtract 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 Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples Solve |15 – 3x| = 6. |15 – 3x| = 6 15 – 3x = 6 or 15 – 3x = –6 The value of 15 – 3x can be 6 or –6 since |6| and |–6| both equal 6. x = 3 or x = 7 Divide each side of both equations by –3.

  4. You try: | 2y - 6 | = 12

  5. Extraneous Solution: a solution of an equation derived from an original equation that is not a solution to the original equation. Extraneous Solution Video

  6. 9 2 |x + 9| = Divide each side by –2. 9 2 9 2 x + 9 = or x + 9 =– Rewrite as two equations. Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5 4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5 4 – 2 |4.5| –5 4 – 2 |–4.5| –5 4 – 2 (4.5) –5 4 – 2 (4.5) –5 –5 = –5 –5 = –5 Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples Solve 4 – 2|x + 9| = –5. 4 – 2|x + 9| = –5 –2|x + 9| = –9 Add –4 to each side. x = –4.5 or x =–13.5 Subtract 9 from each side of both equations.

  7. 3x – 4 = –4x – 1 or 3x – 4 = –(–4x –1) Rewrite as two equations. 7x – 4 = –1 3x – 4 = 4x + 1 Solve each equation. 7x = 3 – x = 5 3 7 x = or x = –5 Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples Solve |3x – 4| = –4x – 1. |3x – 4| = –4x – 1

  8. = / 3 7 3 7 The only solution is –5. is an extraneous solution. –4() – 1 Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples (continued) Check: |3x – 4| = –4x – 1 |3x – 4| = –4x – 1 |3( ) – 4| |3(–5) – 4| (–4(–5) –1) 3 7 19 7 19 7 |– | – |–19| 19 19 7 19 7 19 = 19

  9. 2x < 2 2x > 8 Solve for x. Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples Solve |2x – 5| > 3. Graph the solution. |2x – 5| > 3 2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality. x < 1 or x > 4

  10. < < < < < – – – – – > > > – – – –2|x + 1| + 5 –3 –2|x + 1| –8 Isolate the absolute value expression. Subtract 5 from each side. |x + 1| 4 Divide each side by –2 and reverse the inequality. –4 x + 1 4 Rewrite as a compound inequality. –5 x 3 Solve for x. Absolute Value Equations and Inequalities Lesson 1-5 Additional Examples Solve –2|x + 1| + 5 –3. Graph the solution.

  11. Homework pg 36 3-27 every 3rd

More Related