1. Chapter 3 - Inequalities Algebra I

2. Table of Contents 3.1 - Graphing and Writing Inequalities 3.1 3.2 - Solving Inequalities by Adding or Subtracting 3.2 3.3 - Solving inequalities by multiplying or dividing 3.3 3.4 - Solving Two Step and Multi-Step Inequalities 3.4 3.5 - Solving Inequalities with Variables on Both Sides 3.5 3.6 - Solving Compound Inequalities 3.6 3.7- Solving Absolute-Value Inequalities 3.7

3. 3.1- Graphing and Writing Inequalities Algebra I

4. An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs: ≤ A ≤ B A is less than or equal to B. < A < BA < B A is less than B. > A > B A is greater than B. ≥ A ≥ B A is greater than or equal to B. ≠ A ≠ B A is not equal to B. A solution of an inequality is any value that makes the inequality true. 3.1Algebra 1 (bell work)

5. Describe the solutions of x – 6 ≥ 4 in words. It appears that the solutions of x – 6 ≥ 4 are all real numbers greater than or equal to 10. Describe the solutions of 2p > 8 in words. It appears that the solutions of 2p > 8 are all real numbers greater than 4. 3.1Example 1Identifying Solutions of Inequalities +6 X ≥ 10 2 p > 4

6. 3.1 Do not Copy

7. Graph each inequality A. m ≥ 0 1 – 23 3 B. t < 5(–1 + 3) t < 5(–1 + 3) t < 5(2) t < 10 – 4 – 2 0 24681012 – 6 – 8 3.1Example 2Graphing Inequalities

8. Write the inequality shown by each graph x < 2x ≥ –0.5 3.1Example 3Writing an Inequality from a Graph

9. Math Joke Q: Why did the Moore family name their son Lester? A: So he could be called either ‘Moore’ or ‘les’ 3.1

10. Ray’s dad told him not to turn on the air conditioner unless the temperature is at least 85°F. Define a variable and write an inequality for the temperatures at which Ray can turn on the air conditioner. Graph the solutions. Let t represent the temperatures at which Ray can turn on the air conditioner. 75 80859070 Turn on the AC when temperatureis at least85°F t ≥ 85 t  85 3.1Example 4Application

11. A store’s employees earn at least $8.50 per hour Define a variable and write an inequality for the amount the employees may earn per hour. Graph the solutions. Let w represent an employee ’ s wages. An employee earns at least$8.50 w≥8.50 4681012−202141618 8.5 w ≥ 8.5 3.1

12. HW pg. 173 3.1- – 3-15 (Odd), 16, 17, 18-32 (Even), 33, 42-47, 55, 74-81 – Follow All HW Guidelines or ½ off

13. 3.2 - Solving Inequalities by Adding or Subtracting Algebra I

14. 3.2Algebra 1 (bell work) Just Read

15. Solve the inequality and graph the solutions. x + 12 < 20 –12 x + 0 < 8 x < 8 –10 –8 –6–4 –2 0246810 3.2Example 1Using Addition and Subtraction to Solve Inequalities +5 d + 0 > –2 d > –2 d – 5 > –7 –10 –8 –6–4 –2 0246810 d – 5 > –7

16. Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions. You can check the endpoint and the direction of the inequality symbol. The solutions of x + 9 < 15 are given by x < 6. 3.2

17. Math Joke Q: Why did the parents think their little variable was sick? A: The nurse said he had to be isolated 3.2

18. Sami has a gift card. She has already used $14 of the of the total value, which was $30. Write, solve, and graph an inequality to show how much more she can spend. Amount remaining plus $30. is at most amount used g + 14 ≤ 30 g + 14 ≤ 30 – 14 g + 0 ≤ 16 g ≤ 16 0246810 12 14 16 18 10 3.2Example 2/3Application

19. Mrs. Lawrence wants to buy an antique bracelet at an auction. She is willing to bid no more than $550. So far, the highest bid is $475. Write and solve an inequality to determine the amount Mrs. Lawrence can add to the bid. Check your answer. Let x represent the amount Mrs. Lawrence can add to the bid. $475plus amount can add is at most $550. x + 475 ≤ 550 475 + x ≤ 550 –475 – 475 x ≤ 75 0 + x ≤ 75 3.2

20. Josh wants to try to break the school bench press record of 282 pounds. He currently can bench press 250 pounds. Write and solve an inequality to determine how many more pounds Josh needs to lift to break the school record. Check your answer. Let p represent the number of additional pounds Josh needs to lift. 250 pounds plus additional pounds is greater than 282 pounds. 250 + p>282 250 + p > 282 –250 p > 32 3.2

21. HW pg. 179 3.2- – 1-15, 25-31, 47-56 – Ch: 35 – Follow All HW Guidelines or ½ off

22. 3.3 - Solving Inequalities by Multiplying or Dividing Algebra I

23. 3.3Algebra 1 (bell work) Just Read

24. Solve the inequality and graph the solutions. 7x > –42 > 1x > –6 x > –6 –10 –8 –6–4 –2 0246810 3.3Example 1Multiplying or Dividing by a Positive Number 3(2.4) ≤ 3 m ≥ 7.2) 0246810 12 14 16 18 20

25. r < 16 0246810 12 14 16 18 20 Solve the inequality and graph the solutions 3.3

26. If you multiply or divide both sides of an inequality by a negative number, the resulting inequality is not a true statement. You need to reverse the inequality symbol to make the statement true. 3.3

27. Math Joke Q: What did the teacher do to prepare for class? A: She made a “less-than” plan (lesson plan) 3.3

28. Solve the inequality and graph the solutions. –12x > 84 x < –7 –10 –8–8 –6–6–4–4 –2–2 0246 –12–14 –7 3.3Example 2Multiplying or Dividing by a Negative Number 16182022241014262830 12 x  24)

29. Solve each inequality and graph the solutions a. 10 ≥ –x –1(10) ≤ –1(–x) –10 ≤ x b. 4.25 > –0.25h –17 < h –20 –16 –12–8 –4 0481216 20 – 17 –10 –8 –6–4 –2 0246810 3.3

30. $4.30 times number of tubes is at most $20.00. 4.30 p ≤ 20.00 Jill has a $20 gift card to an art supply store where 4 oz tubes of paint are $4.30 each after tax. What are the possible numbers of tubes that Jill can buy? Let p represent the number of tubes of paint that Jill can buy. 4.30p ≤ 20.00 p ≤ 4.65 Since Jill can buy only whole numbers of tubes, she can buy 0, 1, 2, 3, or 4 tubes of paint. 3.3Example 3Application

31. A pitcher holds 128 ounces of juice. What are the possible numbers of 10-ounce servings that one pitcher can fill? 10 oz times number of servings is at most 128 oz 10 x ≤ 128 Let x represent the number of servings of juice the pitcher can contain. 10x ≤ 128 x ≤ 12.8 The pitcher can fill 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12 servings. 3.3

32. HW pg. 185 3.3- – 19-41 (Odd), 17, 42, 51-54, 66, 78-87 – Ch: 62 – Follow All HW Guidelines or ½ off

33. 3.4 - Solving Two-Step and Multi- Step Inequalities Algebra I

34. Solve the inequality and graph the solutions. 45 + 2b > 61 –45 2b > 16 b > 8 0246810 12 14 16 18 20 3.4Example 1Solving Multi-Step Inequalities –12 ≥ 3x + 6 – 6 –18 ≥ 3x –6 ≥ x –10 –8 –6–4 –2 0246810

35. Solve the inequality and graph the solutions. x < –11 –5 x + 5 < –6 –20 –12–8–4 –16 0 –11 3.4Example 2Simplifying Before Solving Inequalities –4(2 – x) ≤ 8 −4(2 – x) ≤ 8 −4(2) − 4(−x) ≤ 8 –8 + 4x ≤ 8 +8 4x ≤ 16 x ≤ 4 –10 –8 –6–4 –2 02468 10

36. Solve the inequality and graph the solutions. 4f + 3 > 2 –3 4f > –1 0 3.4

37. Math Joke Q: What did the doctor say to the multi-step inequality? A: I can solve your problems with a few operations 3.4

38. To rent a certain vehicle, Rent-A-Ride charges $55.00 per day with unlimited miles. The cost of renting a similar vehicle at We Got Wheels is $38.00 per day plus $0.20 per mile. For what number of miles in the cost at Rent-A-Ride less than the cost at We Got Wheels? Let m represent the number of miles. The cost for Rent-A-Ride should be less than that of We Got Wheels. Cost at Rent-A- Ride must be less than daily cost at We Got Wheels plus $0.20 per mile times # of miles. 55 < 38 +0.20  m 3.4Example 3Application

39. 85 < m Rent-A-Ride costs less when the number of miles is more than 85. 55 < 38 + 0.20m –38 55 < 38 + 0.20m 17 < 0.20m 3.4

40. The average of Jim ’ s two test scores must be at least 90 to make an A in the class. Jim got a 95 on his first test. What grades can Jim get on his second test to make an A in the class? Let x represent the test score needed. The average score is the sum of each score divided by 2. First test score plus second test score divided by number of scores is greater than or equal to total score (95 + x) x)  2 ≥ 90 3.4

41. The score on the second test must be 85 or higher. 95 + x ≥ 180 –95 x ≥ 85 3.4

42. HW pg.193 3.4- – 15-37 (Odd), 44, 49,51-54, 59, 77-86 – Ch: 61 – Follow All HW Guidelines or ½ off

43. 3.5 - Solving Inequalities with Variables on Both Sides Algebra I

44. Solve the inequality and graph the solutions. y ≤ 4y + 18 –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y y  –6 –10 –8 –6–4 –2 0246810 3.5Example 1Solving Inequalities with Variables on Both Sides 4m – 3 < 2m + 6 –2m – 2m 2m – 3 < + 6 + 3 2m < 9 4 5 6

45. Solve the inequality and graph the solutions 5t + 1 < –2t – 6 +2t 7t + 1 < –6 – 1 < –1 7t < –7 7 t < –1 –5 –4 –3–2 –1 01234 5 3.5

46. The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding. How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean? Let w be the number of windows. 312 + 12 w < 36 w 312 < 24w 13 < w The Home Cleaning Company is less expensive for houses with more than 13 windows. Home Cleaning Company siding charge plus $12 per window # of windows is less than Power Clean cost per window # of windows. times 3.5Example 2Application

47. Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 2(k – 3) > 3 + 3k 2k + 2(–3) > 3 + 3k 2k – 6 > 3 + 3k –2k – 2k –6 > 3 + k –3 –9 > k 3.5Example 3Simplifying Each Side Before Solving

48. 3.5Just Read

49. Math Joke Q: What did Miss Manners say to the inequality symbol? A: It’s not polite to point 3.5

50. Solve the inequality. 2x – 7 ≤ 5 + 2x –2x –7 ≤ 5 The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions. 3.5Example 4All Real Numbers as Solutions or No Solutions 2(3y – 2) – 4 ≥ 3(2y + 7) 2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7) 6y – 4 – 4 ≥ 6y + 21 6y – 8 ≥ 6y + 21 –6y –8 ≥ 21  No values of y make the inequality true. There are no solutions.

51. 4(y – 1) ≥ 4y + 2 4(y) + 4(–1) ≥ 4y + 2 4y – 4 ≥ 4y + 2 Solve the inequality. –4y –4 ≥ 2  No values of y make the inequality true. There are no solutions. 3.5Optional

52. HW pg. 199 3.5- – 1-17, 26, 45-49 – Ch: 50-56 – Follow All HW Guidelines or ½ off

53. 3.6 - Solving Compound Inequalities Algebra I

54. 3.6Algebra 1 (bell work) Copy Boxed Parts Pg. 204

55. The pH level of a popular shampoo is between 6.0 and 6.5 inclusive. Write a compound inequality to show the pH levels of this shampoo. Graph the solutions. Let p be the pH level of the shampoo. 6.0is less than or equal to pH level is less than or equal to 6.5 6.0 ≤ p ≤ 6.5 5.9 6.16.26.3 6.0 6.4 6.5 3.6Example 1Application

56. The free chlorine in a pool should be between 1.0 and 3.0 parts per million inclusive. Write a compound inequality to show the levels that are within this range. Graph the solutions. Let c be the chlorine level of the pool. 1.0is less than or equal to chlorine is less than or equal to 3.0 1.0 ≤ c ≤ 3.0 0 23 4 1 5 6 3.6

57. In this diagram, oval A represents some integer solutions of x 0. The overlapping region represents numbers that belong in both ovals. Those numbers are solutions of both x 0. 3.6

58. You can graph the solutions of a compound inequality involving AND by using the idea of an overlapping region. The overlapping region is called the intersection and shows the numbers that are solutions of both inequalities. 3.6

59. Solve the compound inequality and graph the solutions. –5 < x + 1 < 2 –1 – 1 – 1 –6 < x < 1 –10 –8 –6–4 –2 0246810 3.6Example 2Solving Compound Inequalities Involving AND 8 < 3x – 1 ≤ 11 +1 +1 +1 9 < 3x ≤ 12 3 < x ≤ 4

60. In this diagram, circle A represents some integer solutions of x 10. The combined shaded regions represent numbers that are solutions of either x 10. 3.6

61. Math Joke Q: How does a math teacher get a compound fracture? A: She breaks her hAND 3.6

62. You can graph the solutions of a compound inequality involving OR by using the idea of combining regions. The combine regions are called the union and show the numbers that are solutions of either inequality. 3.6Day 2

63. Solve the inequality and graph the solutions. 8 + t ≥ 7 OR 8 + t < 2 –8 –8 –8 −8 t ≥ –1 OR t < –6 –10 –8 –6–4 –2 0246810 3.6Example 3Solving Compound Inequalities Involving OR 4x ≤ 20 OR 3x > 21 x ≤ 5 OR x > 7 0246810 –10 –8 –6–4 –2

64. Every solution of a compound inequality involving AND must be a solution of both parts of the compound inequality. If no numbers are solutions of both simple inequalities, then the compound inequality has no solutions. The solutions of a compound inequality involving OR are not always two separate sets of numbers. There may be numbers that are solutions of both parts of the compound inequality. 3-6

65. Write the compound inequality shown by the graph. The shaded portion of the graph is not between two values, so the compound inequality involves OR. The compound inequality is x ≤ –8 OR x > 0. 3.6Example 4Writing a Compound Inequality from a Graph The shaded portion of the graph is between the values –2 and 5, so the compound inequality involves AND. The compound inequality is m > –2 AND m < 5 (or -2 < m < 5).

66. The shaded portion of the graph is between the values –9 and –2, so the compound inequality involves AND. The compound inequality is – 9 < x AND x < –2 (or –9 < x < –2). Write the compound inequality shown by the graph. 3.6 The shaded portion of the graph is not between two values, so the compound inequality involves OR. The compound inequality is x ≤ –3 OR x ≥ 2.

67. HW pg. 208 3.6- – Day 1: 1-6, 15-19, 57-65 – Day 2: 7-14, 24-28, 34, 44 – Ch: 29 – Follow All HW Guidelines or ½ off

68. 3.7- Solving Absolute Value Inequalities Algebra I

69. 3.7

70. Solve the inequality and graph the solutions. Then write the solutions as a compound inequality. |x| + 2 ≤ 6 –2 |x| ≤ 4 x ≥ –4 AND x ≤ 4 –5 –4 –3–2 –1 012345 4 units –4 ≤ x ≤ 4 3.7

71. |x| – 5 < –4 +5 |x| < 1 –1 < x AND x < 1 –1 < x < 1 –2 –1 012 unit 1 1 Solve the inequality and graph the solutions. Then write the solutions as a compound inequality. 3.7

73. Math Joke Q: What does an absolute-value expression work on when it goes to the gym? A: Its “abs”! 3.7

74. Solve the inequality and graph the solutions. Then write the solutions as a compound inequality. |x| + 2 > 7 – 2 –2 |x| > 5 x 5 –10 –8 –6–4 –2 0246810 5 units 3.7

75. Solve the inequality and graph the solutions. Then write the solutions as a compound inequality. |x + 3| – 5 > 9 + 5 +5 |x + 3| > 14 –16 –12 –8–404 81216 x + 3 14 14 units 3.7

76. |x| – 7 > –1 +7 |x| > 6 –10 –8 –6–4 –2 0246810 x 6 Solve the inequality and graph the solutions. Then write the solutions as a compound inequality. 6 units 3.7

77. HW pg. 215 3.7- – 1-12, 14-22, 61-64 – Follow All HW Guidelines or ½ off