1. Chapter 1.7 Solve Absolute Value Equations and Inequalities Analyze Situations using algebraic symbols; Use models to understand relationships

2. Chapter 1.7 Solve Absolute Value Equations and Inequalities

5. EXAMPLE 2 Solve an absolute value equation | 5x – 10 | = 45 5x – 10 = 45 or 5x – 10 = –45 5x = 55 or 5x = –35 x = 11 or x = –7 Write original equation. Expression can equal 45 or –45. Add 10 to each side. Divide each side by 5. Solve |5x – 10 | = 45. SOLUTION

6. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 4. |3x – 2| = 13 for Examples 1, 2 and 3 ANSWER

7. Chapter 1.7 Solve Absolute Value Equations and Inequalities

8. EXAMPLE 4 Solve an inequality of the form |ax + b| > c Solve |4x + 5| > 13. Then graph the solution. SOLUTION First Inequality Second Inequality 4x + 5 < –134x + 5 > 13 4x < –184x > 8 x < – 9 2 x > 2 Write inequalities. Subtract 5 from each side. Divide each side by 4. The absolute value inequality is equivalent to 4x +5 13.

9. EXAMPLE 4 ANSWER Solve an inequality of the form |ax + b| > c The solutions are all real numbers less than or greater than 2. The graph is shown below. – 9 2

10. GUIDED PRACTICE for Example 4 Solve the inequality. Then graph the solution. 7. |x + 4| ≥ 6 x 2 The graph is shown below. ANSWER

11. GUIDED PRACTICE for Example 4 Solve the inequality. Then graph the solution. 8. |2x –7|>1 ANSWER x 4 The graph is shown below.

12. GUIDED PRACTICE for Example 4 Solve the inequality. Then graph the solution. 9. |3x + 5| ≥ 10 ANSWER x 1 2 3 The graph is shown below.

13. EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball. Baseball SOLUTION Write a verbal model. Then write an inequality. STEP 1

14. EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c STEP 2Solve the inequality. Write inequality. Write equivalent compound inequality. Add 5.125 to each expression. |w – 5.125| ≤ 0.125 – 0.125 ≤ w – 5.125 ≤ 0.125 5 ≤ w ≤ 5.25 So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below. ANSWER

15. GUIDED PRACTICE for Examples 5 and 6 Solve the inequality. Then graph the solution. 10. |x + 2| < 6 The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below. ANSWER –8 < x < 4

16. GUIDED PRACTICE for Examples 5 and 6 Solve the inequality. Then graph the solution. 11. |2x + 1| ≤ 9 The solutions are all real numbers less than –5 or greater than 4. The graph is shown below. ANSWER –5 ≤ x ≤ 4

17. GUIDED PRACTICE for Examples 5 and 6 12. |7 – x| ≤ 4 Solve the inequality. Then graph the solution. 3 ≤ x ≤ 11 ANSWER The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.

18. Solve |x – 5| = 7. Graph the solution. SOLUTION | x – 5 | = 7 x – 5 = – 7 or x – 5 = 7 x = 5 – 7 or x = 5 + 7 x = –2 or x = 12 Write original equation. Write equivalent equations. Solve for x. Simplify. The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.

19. EXAMPLE 3 | 2x + 12 | = 4x 2x + 12 = 4x or 2x + 12 = – 4x 12 = 2x or 12 = –6x 6 = x or –2 = x Write original equation. Expression can equal 4x or – 4 x Add –2x to each side. Solve |2x + 12 | = 4x. Check for extraneous solutions. SOLUTION Solve for x. Check for extraneous solutions

20. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 2. |x – 3| = 10 for Examples 1, 2 and 3 The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below. ANSWER – 3 – 4 – 2 – 1 0 12345 6 7 – 5 – 6 – 78 9 10 11 12 13 10

21. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 3. |x + 2| = 7 for Examples 1, 2 and 3 The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line. ANSWER

22. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 5. |2x + 5| = 3x for Examples 1, 2 and 3 The solution of is 5. Reject 1 because it is an extraneous solution. ANSWER

23. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. 6. |4x – 1| = 2x + 9 for Examples 1, 2 and 3 ANSWER The solutions are – and 5. 3 1 1

24. Chapter 1.7 Solve Absolute Value Equations and Inequalities Homework Pg. 55 22-38 evens, 44-52 evens, 66-69