Presentation Transcript

1. Warm Up Solve. 3x + 2 = 8 2. 3. -2k + 7 = -3
2. Objectives Identify solutions of inequalities with one variable. Write and graph inequalities with one variable.
3. < > ≥ ≤ ≠ A<B A >B A ≤ B A ≥B A ≠ B A is greater than or equal to B. A is less than or equal toB. Ais less thanB. A is greater thanB. A is not equal toB. An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs: A solution of an inequality is any value that makes the inequality true.
4. 10.1 –3 0 9.9 10 12 x –9 4.1 –6 3.9 4 6 x – 6 ? ? ? ? –6 4 –9 4 3.9 4 ≥ ≥ ≥ x – 6 ≥ 4 Yes Yes Yes No No No Solution? ? ? ? 6 4 4 4 4.1 4 ≥ ≥ ≥ Example 1: Identifying Solutions of Inequalities Describe the solutions of x – 6 ≥ 4 in words. When the value of x is a number less than 10, the value of x – 6 is less than 4. When the value of x is 10, the value of x – 6 is equal to 4. When the value of x is a number greater than 10, the value of x – 6 is greater than 4. It appears that the solutions of x – 6 ≥ 4 are all real numbers greater than or equal to 10.
5. An inequality like 3 + x < 9 has too many solutions to list. You can use a graph on a number line to show all the solutions. The solutions are shaded and an arrow shows that the solutions continue past those shown on the graph. To show that an endpoint is a solution, draw a solid circle at the number. To show an endpoint is not a solution, draw an empty circle.
6. 0 2 3 3 – t < 5(–1 + 3) 1 t < 5(2) t < 10 0 –8 –6 –4 –2 2 4 6 8 10 12 Example 2: Graphing Inequalities Graph each inequality. A. m ≥ B. t < 5(–1 + 3)
7. Example 3: Writing an Inequality from a Graph Write the inequality shown by each graph. x < 2 Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2 means that 2 is not a solution, so use <. x ≥ –0.5 Use any variable. The arrow points to the right, so use either > or ≥. The solid circle at –0.5 means that –0.5 is a solution, so use ≥.
8. Reading Math “No more than” means “less than or equal to.” “At least” means “greater than or equal to”.
9. Turn on the AC when temperature is at least 85°F t 85 ≥ 75 70 80 85 90 Example 4:Application 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. Draw a solid circle at 85. Shade all numbers greater than 85 and draw an arrow pointing to the right. t 85
10. An employee earns at least $8.50 w ≥ 8.50 8.5 −2 0 2 4 6 8 10 12 14 16 18 Check It Out! Example 4 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. w ≥ 8.5
11. –4.75 –5 –4.5 Warm-Up 1. Describe the solutions of 7 < x + 4. 2. Graph h ≥ –4.75 Write the inequality shown by each graph. 3. 4. 5. A cell phone plan offers free minutes for no more than 250 minutes per month. Define a variable and write an inequality for the possible number of free minutes. Graph the solution.
12. Objectives Solve one-step inequalities by using addition. Solve one-step inequalities by using subtraction.
13. –12 –12 –8 –2 –10 –6 –4 0 2 4 6 8 10 Example 1A: Using Addition and Subtraction to Solve Inequalities Solve the inequality and graph the solutions. x + 12 < 20 x + 12 < 20 x + 0 < 8 x < 8
14. d – 5 > –7 +5 +5 d + 0 > –2 d > –2 –8 –2 –10 –6 –4 0 2 4 6 8 10 Example 1B: Using Addition and Subtraction to Solve Inequalities Solve the inequality and graph the solutions. d – 5 > –7
15. +0.3 +0.3 1.2 ≥ n –0 1.2 ≥ n 1.2  Example 1C: Using Addition and Subtraction to Solve Inequalities Solve the inequality and graph the solutions. 0.9 ≥ n – 0.3 0.9 ≥ n – 0.3 1 0 2
16. Example 2: Problem-Solving Application 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.
17. Make a Plan Amount remaining is at most plus amount used $30. g + 14 ≤ 30 Example 2 Continued Write an inequality. Let g represent the remaining amount of money Sami can spend. g + 14 ≤ 30
18. Solve – 14 – 14 0 2 4 6 8 10 14 10 12 18 16 Example 2 Continued g + 14 ≤ 30 g + 0 ≤ 16 g ≤ 16
19. Check It Out! Example 2 The Recommended Daily Allowance (RDA) of iron for a female in Sarah’s age group (14-18 years) is 15 mg per day. Sarah has consumed 11 mg of iron today. Write and solve an inequality to show how many more milligrams of iron Sarah can consume without exceeding RDA.
20. Make a Plan Write an inequality. Let x represent the amount of iron Sarah needs to consume. amount needed is at most Amount taken plus 15 mg 11 + x  15 Check It Out! Example 2 Continued 11 + x  15
21. Solve –11 –11 0 1 2 3 4 5 7 10 6 9 8 Check It Out! Example 2 Continued 11 + x 15 x  4 x 4. Sarah can consume 4 mg or less of iron without exceeding the RDA.
22. Warm-Up Solve each inequality and graph the solutions. 1. 13 < x + 7 2. –6 + h ≥ 15 x > 6 h ≥ 21 3. 6.7 + y ≤ –2.1 y ≤ –8.8 4. A certain restaurant has room for 120 customers. On one night, there are 72 customers dining. Write and solve an inequality to show how many more people can eat at the restaurant.
23. Objectives Solve one-step inequalities by using multiplication. Solve one-step inequalities by using division.
24. Remember, solving inequalities is similar to solving equations. To solve an inequality that contains multiplication or division, undo the operation by dividing or multiplying both sides of the inequality by the same number. The following rules show the properties of inequality for multiplying or dividing by a positive number. The rules for multiplying or dividing by a negative number appear later in this lesson.
25. 7x > –42 > –8 –2 –10 –6 –4 0 2 4 6 8 10 Example 1A: Multiplying or Dividing by a Positive Number Solve the inequality and graph the solutions. 7x > –42 1x > –6 x > –6
26. 3(2.4) ≤ 3 0 2 4 6 8 10 14 20 12 18 16 Example 1B: Multiplying or Dividing by a Positive Number Solve the inequality and graph the solutions. 7.2 ≤ m (or m ≥ 7.2)
27. 0 2 4 6 8 10 14 20 12 18 16 Example 1C: Multiplying or Dividing by a Positive Number Solve the inequality and graph the solutions. r < 16
28. 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.
29. Caution! Do not change the direction of the inequality symbol just because you see a negative sign. For example, you do not change the symbol when solving 4x < –24.
30. –7 –14 –12 –8 –2 –10 –6 –4 0 2 4 6 Example 2A: Multiplying or Dividing by a Negative Number Solve the inequality and graph the solutions. –12x > 84 x < –7
31. 10 14 16 18 20 22 24 26 28 30 12 Example 2B: Multiplying or Dividing by a Negative Number Solve the inequality and graph the solutions. 24  x (or x  24)
32. number of tubes is at most times $4.30 $20.00. 20.00 4.30 ≤ p • Example 3:Application 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.
33. Example 3 Continued 4.30p ≤ 20.00 Since p is multiplied by 4.30, divide both sides by 4.30. The symbol does not change. p ≤ 4.65… Since Jill can buy only whole numbers of tubes, she can buy 0, 1, 2, 3, or 4 tubes of paint.
34. Warm-Up Solve each inequality and graph the solutions. 1. 8x < –24 x < –3 2. –5x ≥30 x ≤ –6 4. 3. x > 20 x ≥ 6 5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy? 0, 1, 2, 3, 4, 5, 6, or 7 shirts
35. Warm-Up Solve each inequality and graph the solutions. 1. 8x < –24 x < –3 2. –5x ≥30 x ≤ –6 4. 3. x > 20 x ≥ 6 5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy? 0, 1, 2, 3, 4, 5, 6, or 7 shirts
36. Objective Solve inequalities that contain more than one operation.
37. Inequalities that contain more than one operation require more than one step to solve. Use inverse operations to undo the operations in the inequality one at a time.
38. 45 + 2b > 61 –45 –45 b > 8 0 2 4 6 8 10 14 20 12 18 16 Example 1A: Solving Multi-Step Inequalities Solve the inequality and graph the solutions. 45 + 2b > 61 2b > 16
39. 8 – 3y ≥ 29 –8 –8 –7 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 1B: Solving Multi-Step Inequalities Solve the inequality and graph the solutions. 8 – 3y ≥ 29 –3y ≥21 y ≤ –7
40. –1 –1 –10 –20 –16 –12 –8 –4 0 Check It Out! Example 1c Solve the inequality and graph the solutions. 1 – 2n ≥21 –2n ≥ 20 n ≤ –10
41. To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides by using the order of operations, combining like terms, or using the Distributive Property.
42. 10 –8 –10 –6 –4 0 2 4 6 8 –2 –3 Example 2A: Simplifying Before Solving Inequalities Solve the inequality and graph the solutions. 2 – (–10) > –4t 12 > –4t –3 < t (or t > –3)
43. –8 + 4x ≤ 8 +8 +8 10 –8 –10 –6 –4 0 2 4 6 8 –2 Example 2B: Simplifying Before Solving Inequalities Solve the inequality and graph the solutions. –4(2 – x) ≤ 8 −4(2 – x) ≤ 8 −4(2) − 4(−x) ≤ 4x ≤16 x ≤ 4
44. – 11 –11 10 –8 –10 –6 –4 0 2 4 6 8 –2 Check It Out! Example 2b Solve the inequality and graph the solutions. 3 + 2(x + 4) > 3 3 + 2(x + 4) > 3 3 + 2x + 8 > 3 2x + 11 > 3 2x > –8 x > –4
45. Check It Out! Example 2c Solve the inequality and graph the solutions. 5 < 3x –2 +2 + 2 7 < 3x
46. 2 4 6 8 0 10 Check It Out! Example 2c Continued Solve the inequality and graph the solutions. 7 < 3x
47. daily cost at We Got Wheels Cost at Rent-A-Ride must be less than $0.20 per mile # of miles. plus times 55 < 38 m + 0.20  Example 3: Application 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 is 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.
48. 55 < 38 + 0.20m –38 –38 Example 3 Continued 55 < 38 + 0.20m Since 38 is added to 0.20m, subtract 8 from both sides to undo the addition. 17 < 0.20m Since m is multiplied by 0.20, divide both sides by 0.20 to undo the multiplication. 85 < m Rent-A-Ride costs less when the number of miles is more than 85.
49. is greater than or equal to First test score second test score divided by number of scores total score plus 90 ≥  (95 x) 2 + Check It Out! Example 3 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.
50. –95 –95 Check It Out! Example 3 Continued Since 95 + x is divided by 2, multiply both sides by 2 to undo the division. 95 + x ≥ 180 Since 95 is added to x, subtract 95 from both sides to undo the addition. x ≥ 85 The score on the second test must be 85 or higher.
51. Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. 13 – 2x ≥ 21 x ≤–4 2. –11 + 2 < 3p p > –3 t > 7 3. 23 < –2(3 –t) 4.
52. Lesson Quiz: Part II 5. A video store has two movie rental plans. Plan A includes a $25 membership fee plus $1.25 for each movie rental. Plan B costs $40 for unlimited movie rentals. For what number of movie rentals is plan B less than plan A? more than 12 movies
53. –4 –3 –2 –1 –5 –6 0 Warm Up Solve each equation. 1. 2x = 7x + 15 2. x = –3 3y – 21 = 4 – 2y y = 5 z = –1 3. 2(3z + 1) = –2(z + 3) 4. 3(p –1) = 3p + 2 no solution b < –3 5. Solve and graph 5(2 –b) > 52.
54. Objective Solve inequalities that contain variable terms on both sides.
55. y ≤ 4y + 18 –y –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 1A: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. y ≤ 4y + 18 –6 ≤ y (or y –6)
56. –2m –2m 2m – 3 < + 6 + 3 + 3 2m < 9 4 5 6 Example 1B: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. 4m – 3 < 2m + 6
57. Example 2: Business Application 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.
58. Home Cleaning Company siding charge Power Clean cost per window is less than # of windows. # of windows times $12 per window plus times – 12w –12w Example 2 Continued 312 + 12 • w < 36 • w 312 + 12w < 36w To collect the variable terms, subtract 12w from both sides. 312 < 24w Since w is multiplied by 24, divide both sides by 24 to undo the multiplication. 13 < w The Home Cleaning Company is less expensive for houses with more than 13 windows.
59. $0.10 per flyer Print and More’s cost per flyer A-Plus Advertising fee of $24 # of flyers. # of flyers is less than times times plus Check It Out! Example 2 A-Plus Advertising charges a fee of $24 plus $0.10 per flyer to print and deliver flyers. Print and More charges $0.25 per flyer. For how many flyers is the cost at A-Plus Advertising less than the cost of Print and More? Let f represent the number of flyers printed. 24 + 0.10 • f < 0.25 • f
60. –0.10f –0.10f Check It Out! Example 2 Continued 24 + 0.10f < 0.25f To collect the variable terms, subtract 0.10f from both sides. 24 < 0.15f Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication. 160 < f More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company.
61. You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.
62. +6 +6 –6 –4 –2 0 2 4 + 5r +5r Check It Out! Example 3a Solve the inequality and graph the solutions. 5(2 – r) ≥ 3(r – 2) 16 ≥ 8r 5(2 – r) ≥ 3(r – 2) 5(2) –5(r) ≥ 3(r) + 3(–2) 10 – 5r ≥ 3r – 6 2 ≥ r 16 − 5r ≥ 3r 16 ≥ 8r
63. There are special cases of inequalities called identities and contradictions.
64. –2x –2x Example 4A: Identities and Contradictions Solve the inequality. 2x – 7 ≤ 5 + 2x 2x – 7 ≤ 5 + 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.
65. –6y –6y Example 4B: Identities and Contradictions Solve the inequality. 2(3y – 2) – 4 ≥ 3(2y + 7) 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  –8 ≥ 21 No values of y make the inequality true. There are no solutions.
66. Warm-Up Solve each inequality and graph the solutions. 1. t < 5t + 24 2. 5x – 9 ≤ 4.1x – 81 3. 4b + 4(1 – b) > b – 9 Solve each inequality. 4. 2y – 2 ≥ 2(y + 7) 5. 2(–6r – 5) < –3(4r + 2)
67. Objectives Solve compound inequalities with one variable. Graph solution sets of compound inequalities with one variable.
68. The inequalities you have seen so far are simple inequalities. When two simple inequalities are combined into one statement by the words AND or OR, the result is called a compound inequality.
69. 6.0 is less than or equal to pH level 6.5 is less than or equal to 5.9 6.0 6.1 6.2 6.3 6.5 6.4 Example 1: Chemistry Application 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. 6.0 ≤ p ≤6.5 6.0 ≤ p ≤ 6.5
70. In this diagram, oval A represents some integer solutions of x < 10 and oval B represents some integer solutions of x > 0. The overlapping region represents numbers that belong in both ovals. Those numbers are solutions of both x < 10 andx > 0.
71. 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.
72. –1 – 1 – 1 Example 2A: Solving Compound Inequalities Involving AND Solve the compound inequality and graph the solutions. –5 < x + 1 < 2 –5 < x + 1 < 2 –6 < x < 1 Graph –6 < x. Graph x < 1. Graph the intersection by finding where the two graphs overlap. –8 –2 –10 –6 –4 0 2 4 6 8 10
73. 8 < 3x – 1 ≤ 11 +1 +1 +1 9 < 3x ≤ 12 Example 2B: Solving Compound Inequalities Involving AND Solve the compound inequality and graph the solutions. 8 < 3x – 1 ≤ 11 3 < x ≤ 4
74. Example 2B Continued Graph 3 < x. Graph x ≤ 4. Graph the intersection by finding where the two graphs overlap. –3 –2 0 1 2 3 4 5 –4 –1 –5
75. –9 < x – 10 < –5 +10 +10 +10 1 < x < 5 Check It Out! Example 2a Solve the compound inequality and graph the solutions. –9 < x – 10 < –5 Graph 1 < x. Graph x < 5. Graph the intersection by finding where the two graphs overlap. –3 –2 0 1 2 3 4 5 –5 –4 –1
76. In this diagram, circle A represents some integer solutions of x < 0, and circle B represents some integer solutions of x > 10. The combined shaded regions represent numbers that are solutions of either x < 0 or x >10.
77. > 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.
78. –8 –8 –8 −8 Example 3A: Solving Compound Inequalities Involving OR Solve the inequality and graph the solutions. 8 + t ≥ 7 OR 8 + t < 2 8 + t ≥ 7 OR 8 + t < 2 t ≥ –1 OR t < –6 Graph t ≥ –1. Graph t < –6. Graph the union by combining the regions. –8 –2 –10 –6 –4 0 2 4 6 8 10
79. 4x ≤ 20 OR 3x > 21 x ≤ 5 OR x > 7 Example 3B: Solving Compound Inequalities Involving OR Solve the inequality and graph the solutions. 4x ≤ 20 OR 3x > 21 Graph x ≤ 5. Graph x > 7. Graph the union by combining the regions. 0 2 4 6 8 10 –8 –2 –10 –6 –4
80. –2 –2 –5 –5 Check It Out! Example 3a Solve the compound inequality and graph the solutions. 2 +r < 12 OR r + 5 > 19 2 +r < 12 OR r + 5 > 19 Solve each simple inequality. r < 10 OR r > 14 Graph r < 10. Graph r > 14. Graph the union by combining the regions. 4 –4 0 2 6 8 10 12 14 16 –2
81. Example 4A: Writing a Compound Inequality from a Graph 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. On the left, the graph shows an arrow pointing left, so use either < or ≤. The solid circle at –8 means –8 is a solution so use ≤. x ≤ –8 On the right, the graph shows an arrow pointing right, so use either > or ≥. The empty circle at 0 means that 0 is not a solution, so use >. x > 0 The compound inequality is x ≤ –8 OR x > 0.
82. Example 4B: Writing a Compound Inequality from a Graph Write the compound inequality shown by the graph. The shaded portion of the graph is between the values –2 and 5, so the compound inequality involves AND. The shaded values are on the right of –2, so use > or ≥. The empty circle at –2 means –2 is not a solution, so use >. m > –2 The shaded values are to the left of 5, so use < or ≤. The empty circle at 5 means that 5 is not a solution so use <. m < 5 The compound inequality is m > –2 AND m < 5 (or -2 < m < 5).
83. Lesson Quiz: Part I 1. The target heart rate during exercise for a 15 year-old is between 154 and 174 beats per minute inclusive. Write a compound inequality to show the heart rates that are within the target range. Graph the solutions. 154 ≤ h ≤ 174
84. Lesson Quiz: Part II Solve each compound inequality and graph the solutions. 2. 2 ≤ 2w + 4 ≤ 12 –1 ≤ w ≤ 4 3. 3 + r > −2 OR 3 + r < −7 r > –5 OR r < –10
85. Lesson Quiz: Part III Write the compound inequality shown by each graph. 4. x < −7 OR x ≥ 0 5. −2 ≤ a < 4
86. Warm-Up 1. The target heart rate during exercise for a 15 year-old is between 154 and 174 beats per minute inclusive. Write a compound inequality to show the heart rates that are within the target range. Graph the solutions. Solve each compound inequality and graph the solutions. 2. 2 ≤ 2w + 4 ≤ 12 3. 3 + r > −2 OR 3 + r < −7
87. Objectives Solve inequalities in one variable involving absolute value expressions.
88. When an inequality contains an absolute-value expression, it can be rewritten as a compound inequality. The inequality |x| < 5 describes all real number whose distance from 0 is less than 5 units. The solutions can be written as -5 < x < 5.
89. – 3 – 3 Example: Solving Absolute Value Inequalities involving <. Solve the inequality and graph the solutions. |x| + 3 < 12 |x + 4| ≤ 2 x + 4 ≤ 2 x + 4 ≥ -2 -4 -4 -4 -4 |x| < 9 x ≤ -2 and x ≥ -6
90. The inequality |x| > 5 describes all real number whose distance from 0 is greater than 5 units. The solutions are all numbers less than -5 or greater than 5. The solutions can be written as x < -5 OR x > 5.
91. + 20 +20 - 5 -5 Example: Solving Absolute Value Inequalities involving >. Solve the inequality and graph the solutions. |x| - 20 > -13 |x - 8|+5 ≥ 11 |x – 8| ≥ 6 |x| > 7 x - 8 ≥ 6 x - 8 ≤ -6 x < -7 or x > 7 +8 +8 +8 +8 x ≥ 14 or x ≤ 2