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

2. Absolute Value Equations and Inequalities Algebra2

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 -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 1.ax+b 0 Becomes an “and” problem Changes to: ax+b -c 2.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 -3 7 8

10. Solve & graph.  Get absolute value by itself first.  Becomes an “or” problem -2 3 4 -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. Example 1: ● |2x + 1| > 7 ● 2x + 1 > 7 or 2x + 1 >7 ● 2x + 1 >7 or 2x + 1 <-7 ● x > 3 or x < -4 This is an ‘or’ statement. (Greator). Rewrite. In the 2 nd inequality, reverse the inequality sign and negate the right side value. Solve each inequality. Graph the solution. 3 -4

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

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

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

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

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

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

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