1 / 21

Warm Up Solve.

Warm Up Solve. 1. y + 7 < –11. y < –18. 2. 4 m ≥ –12. m ≥ –3. 3. 5 – 2 x ≤ 17. x ≥ –6. Use interval notation to indicate the graphed numbers. 4. (-2, 3]. (- , 1]. 5. Absolute Value Equations and Inequalities. College Algebra. Absolute Value (of x). Symbol lxl

honorato-pearson
Download Presentation

Warm Up Solve.

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. Warm Up Solve. 1. y + 7 < –11 y< –18 2. 4m ≥ –12 m≥ –3 3. 5 – 2x ≤ 17 x ≥ –6 Use interval notation to indicate the graphed numbers. 4. (-2, 3] (-, 1] 5.

  2. Absolute Value Equations and Inequalities College Algebra

  3. Absolute Value (of x) • Symbol lxl • The distance x is from 0 on the number line. • Always positive • Ex: l-3l = 3 -4 -3 -2 -1 0 1 2

  4. Ex: x = 5 • What are the possible values of x? x = 5 or x = -5

  5. To solve an absolute value equation: ax+b = c, where c > 0 To solve, set up 2 new equations, then solve each equation. ax + b = c or ax + b = -c ** make sure the absolute value is by itself before you split to solve.

  6. Ex: Solve 6x - 3 = 15 6x-3 = 15 or 6x-3 = -15 6x = 18 or 6x = -12 x = 3 or x = -2 * Plug in answers to check your solutions!

  7. Ex: Solve 2x + 7 - 3 = 8 Get the abs. value part by itself first! 2x+7 = 11 Now split into 2 parts. 2x+7 = 11 or 2x+7 = -11 2x = 4 or 2x = -18 x = 2 or x = -9 Check the solutions.

  8. Solving Absolute Value Inequalities • ax+b < c, where c > 0 Becomes an “and” problem Changes to: ax+b < c and ax+b > -c • ax+b > c, where c > 0 Becomes an “or” problem Changes to: ax+b > c or ax+b < -c “less thAND” “greatOR”

  9. Ex: Solve & graph. • Becomes an “and” problem -3 7 8

  10. Solve & graph. • Get absolute value by itself first. • Becomes an “or” problem -2 3 4

  11. Solving an Absolute Value Equation Solve x=7 or x=−2

  12. Solving with less than Solve

  13. Solving with greater than Solve

  14. -4 3 Example 1: This is an ‘or’ statement. (Greator). Rewrite. In the 2nd inequality, reverse the inequality sign and negate the right side value. Solve each inequality. Graph the solution. • |2x + 1| > 7 • 2x + 1 > 7 or 2x + 1 >7 • 2x + 1 >7 or 2x + 1 <-7 • x > 3 or x < -4

  15. 2 8 Example 2: This is an ‘and’ statement. (Less thand). Rewrite. In the 2nd inequality, reverse the inequality sign and negate the right side value. Solve each inequality. Graph the solution. • |x -5|< 3 • x -5< 3 and x -5< 3 • x -5< 3 and x -5> -3 • x < 8 and x > 2 • 2 < x < 8

  16. Solve the equation. This can be read as “the distance from k to –3 is 10.” |–3 + k| = 10 Rewrite the absolute value as a disjunction. –3 + k = 10 or –3 + k = –10 Add 3 to both sides of each equation. k = 13 or k = –7

  17. Solve the equation. Isolate the absolute-value expression. Rewrite the absolute value as a disjunction. Multiply both sides of each equation by 4. x = 16 or x = –16

  18. Solve the inequality. Then graph the solution. |–4q + 2| ≥ 10 Rewrite the absolute value as a disjunction. –4q + 2 ≥ 10 or –4q + 2 ≤ –10 Subtract 2 from both sides of each inequality. –4q ≥ 8 or –4q ≤ –12 Divide both sides of each inequality by –4 and reverse the inequality symbols. q ≤ –2 or q ≥ 3

  19. –3 –2 –1 0 1 2 3 4 5 6 Solve the inequality. Then graph the solution. |3x| + 36 > 12 Isolate the absolute value as a disjunction. |3x| > –24 Rewrite the absolute value as a disjunction. 3x > –24 or 3x < 24 Divide both sides of each inequality by 3. x > –8 or x < 8 The solution is allreal numbers, R. (–∞, ∞)

  20. Solve the compound inequality. Then graph the solution set. Multiply both sides by 3. |2x +7| ≤ 3 Rewrite the absolute value as a conjunction. 2x + 7 ≤ 3 and 2x + 7 ≥ –3 Subtract 7 from both sides of each inequality. 2x ≤ –4 and 2x ≥ –10 Divide both sides of each inequality by 2. x ≤ –2 and x ≥ –5

  21. Solve the compound inequality. Then graph the solution set. Multiply both sides by –2, and reverse the inequality symbol. |p – 2| ≤ –6 Rewrite the absolute value as a conjunction. |p – 2| ≤ –6 and p – 2 ≥ 6 Add 2 to both sides of each inequality. p ≤ –4 and p ≥ 8 Because no real number satisfies both p ≤ –4 andp ≥ 8, there is no solution. The solution set is ø.

More Related