LESSON 1–6

1. LESSON 1–6 Solving Compound and Absolute Value Inequalities

2. A. {x | x > 5} B. {x | x < 5} C. {x | x > 6} D. {x | x < 6} Solve the inequality 3x + 7 > 22. Graph the solution set on a number line. 5-Minute Check 1

3. Solve the inequality . Graph the solution set on a number line. A. {w | w ≤ –9} B. {w | w ≥ –9} C. {w | w ≤ –3} D. {w | w ≥ –3} 5-Minute Check 4

4. Targeted TEKS A2.6(F) Solve absolute value linear inequalities. Mathematical Processes A2.1(E), A2.1(F) TEKS

5. You solved one-step and multi-step inequalities. • Solve compound inequalities. • Solve absolute value inequalities. Then/Now

6. compound inequality • intersection • union Vocabulary

7. Concept

8. Solve an “And” Compound Inequality Solve 10  3y – 2 < 19. Graph the solution set on a number line. Method 1 Solve separately. Write the compound inequality using the word and. Then solve each inequality. 10  3y – 2 and 3y – 2 < 19 12  3y 3y < 21 4  yy < 7 4  y < 7 Example 1

9. Solve an “And” Compound Inequality Method 2 Solve both together. Solve both parts at the same time by adding 2 to each part. Then divide each part by 3. 10  3y – 2 < 19 12  3y < 21 4  y < 7 Example 1

10. y 4 y< 7 4  y< 7 Solve an “And” Compound Inequality Graph the solution set for each inequality and find their intersection. Answer: The solution set is y | 4  y < 7. Example 1

11. A. B. C. D. What is the solution to 11  2x + 5 < 17? Example 1

12. Concept

13. x  4 –x  –4 x + 3 < 2 x < –1 or x< –1 x 4 x < –1 or x 4 Solve an “Or” Compound Inequality Solve x + 3 < 2 or –x –4. Graph the solution set on a number line. Solve each inequality separately. Answer: The solution set is x | x < –1 or x  4. Example 2

14. A. B. C. D. What is the solution to x + 5 < 1 or –2x –6?Graph the solution set on a number line. Example 2

15. Solve Absolute Value Inequalities A. Solve 2 > |d|. Graph the solution set on a number line. 2 > |d| means that the distance between d and 0 on a number line is less than 2 units. To make 2 > |d| true, you must substitute numbers for d that are fewer than 2 units from 0. Notice that the graph of 2 > |d| is the same as the graph of d > –2 and d < 2. All of the numbers between –2 and 2 are less than 2 units from 0. Answer: The solution set is d | –2 < d < 2. Example 3

16. Solve Absolute Value Inequalities B. Solve 3 < |d|. Graph the solution set on a number line. 3 < |d| means that the distance between d and 0 on a number line is greater than 3 units. To make 3 < |d| true, you must substitute values for d that are greater than 3 units from 0. Notice that the graph of 3 < |d| is the same as the graph of d < –3 or d > 3. All of the numbers not between –3 and 3 are greater than 3 units from 0. Answer: The solution set is d | d < –3 or d > 3. Example 3

17. A. B. C. D. A. What is the solution to |x| > 5? Example 3a

18. A.{x | x > 5 or x < –5} B.{x | –5 < x < 5} C.{x | x < 5} D.{x | x > –5} B. What is the solution to |x| < 5? Example 3b

19. Concept

20. Solve a Multi-Step Absolute Value Inequality Solve |2x – 2|  4. Graph the solution set on a number line. |2x – 2|  4 is equivalent to 2x – 2  4 or 2x – 2  –4. Solve each inequality. 2x – 2  4 or 2x – 2  –4 2x  6 2x  –2 x  3 x  –1 Answer: The solution set is x | x  –1 or x  3. Example 4

21. A. B. C. D. What is the solution to |3x – 3| > 9? Graph the solution set on a number line. Example 4

22. The starting salary can differ from the average by as much as $2450. |38,500 – x|  2450 Write and Solve an Absolute Value Inequality A. JOB HUNTINGTo prepare for a job interview, Hinda researches the position’s requirements and pay. She discovers that the average starting salary for the position is $38,500, but her actual starting salary could differ from the average by as much as $2450. Write an absolute value inequality to describe this situation. Let x = the actual starting salary. Answer: |38,500 – x|  2450 Example 5

23. Write and Solve an Absolute Value Inequality B. JOB HUNTINGTo prepare for a job interview, Hinda researches the position’s requirements and pay. She discovers that the average starting salary for the position is $38,500, but her actual starting salary could differ from the average by as much as $2450. Solve the inequality to find the range of Hinda’s starting salary. | 38,500 – x|  2450 Rewrite the absolute value inequality as a compound inequality. Then solve for x. –2450  38,500 – x  2450 –2450 – 38,500  –x  2450 – 38,500 –40,950  –x  –36,050 40,950  x  36,050 Example 5

24. Write and Solve an Absolute Value Inequality Answer: The solution set is x | 36,050  x  40,950.Hinda’s starting salary will fall within $36,050 and $40,950. Example 5

25. A. HEALTHThe average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is an absolute value inequality to describe this situation? A. |4.5 – w|  7 B. |w – 4.5|  7 C. |w – 7|  4.5 D. |7 – w|  4.5 Example 5a

26. B. HEALTHThe average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is the range of birth weights for newborn babies? A. {w | w≤ 11.5} B. {w | w≥ 2.5} C. {w | 2.5 ≤w≤ 11.5} D. {w | 4.5 ≤w≤ 7} Example 5b

27. LESSON 1–6 Solving Compound and Absolute Value Inequalities