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Chapter 10
1. Warm Up Problem of the Day Lesson Presentation 10-1 Solving Two-Step Equations Course 3
2. Warm Up Solve. 1. x + 12 = 35 2. 8 x = 120 3. = 7 4. –34 = y + 56 x = 23 x = 15 y = 63 y = –90 Course 3 10-1 Solving Two-Step Equations y 9
3. Problem of the Day x is an odd integer. If you triple x and then subtract 7, you get a prime number. What is x ? ( Hint: Think about what the prime number must be in order for x to be an odd.) x = 3 Course 3 10-1 Solving Two-Step Equations
5. Sometimes more than one inverse operation is needed to solve an equation. Before solving, ask yourself, “What is being done to the variable, and in what order?” Then work backward to undo the operations. Course 3 10-1 Solving Two-Step Equations
6. The mechanic’s bill to repair Mr. Wong’s car was $650. The mechanic charges $45 an hour for labor, and the parts that were used cost $443. How many hours did the mechanic work on the car? Additional Example 1: Problem Solving Application Course 3 10-1 Solving Two-Step Equations
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8. Think: First the variable is multiplied by 45, and then 443 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 443 from both sides of the equation, and then divide both sides of the new equation by 45. Additional Example 1 Continued 2 Make a Plan Course 3 10-1 Solving Two-Step Equations
9. 650 = 443 + 45 h – 443 –443 Subtract to undo the addition. 207 = 45 h 4.6 = h The mechanic worked for 4.6 hours on Mr. Wong’s car. Additional Example 1 Continued Divide to undo multiplication. Solve 3 Course 3 10-1 Solving Two-Step Equations 207 45 h 45 45 =
10. If the mechanic worked 4.6 hours, the labor would be $45(4.6) = $207. The sum of the parts and the labor would be $443 + $207 = $650. Additional Example 1 Continued Look Back 4 Course 3 10-1 Solving Two-Step Equations
11. The mechanic’s bill to repair your car was $850. The mechanic charges $35 an hour for labor, and the parts that were used cost $275. How many hours did the mechanic work on your car? Try This : Example 1 Course 3 10-1 Solving Two-Step Equations
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13. Think: First the variable is multiplied by 35, and then 275 is added to the result. Work backward to solve the equation. Undo the operations in reverse order: First subtract 275 from both sides of the equation, and then divide both sides of the new equation by 35. Try This : Example 1 Continued 2 Make a Plan Course 3 10-1 Solving Two-Step Equations
14. 850 = 275 + 35 h – 275 –275 Subtract to undo the addition. 575 = 35 h 16.4 h The mechanic worked for about 16.4 hours on your car. Try This : Example 1 Continued Divide to undo multiplication. Solve 3 Course 3 10-1 Solving Two-Step Equations 575 35 h 35 35 =
15. If the mechanic worked 16.4 hours, the labor would be $35(16.4) = $574. The sum of the parts and the labor would be $275 + $574 = $849. Try This : Example 1 Continued Look Back 4 Course 3 10-1 Solving Two-Step Equations
16. Additional Example 2A: Solving Two-Step Equations A. + 7 = 22 Solve. Think: First the variable is divided by 3, and then 7 is added . To isolate the variable, subtract 7, and then multiply by 3 . – 7 – 7 Subtract to undo addition. n = 45 n 3 + 7 = 22 n 3 Multiply to undo division. 3 = 3 15 n 3 Course 3 10-1 Solving Two-Step Equations n 3 = 15
17. Additional Example 2A Continued Substitute 45 into the original equation. Check + 7 = 22 n 3 ? + 7 = 22 45 3 15 + 7 = 22 ? Course 3 10-1 Solving Two-Step Equations
18. Additional Example 2B: Solving Two-Step Equations B. 2.7 = –1.3 m + 6.6 Think: First the variable is multiplied by –1.3, and then 6.6 is added . To isolate the variable, subtract 6.6, and then divide by –1.3 . 2.7 = –1.3 m + 6.6 – 6.6 –6.6 Subtract to undo addition. – 3.9 = –1.3 m Divide to undo multiplication. 3 = m – 1.3 –1.3 – 3.9 = –1.3 m Course 3 10-1 Solving Two-Step Equations
19. Additional Example 2C: Solving Two-Step Equations Think: First 4 is subtracted from the variable, and then the result is divided by 3 . To isolate the variable, multiply by 3, and then add 4 . y – 4 = 27 + 4 + 4 Add to undo subtraction. y = 31 C. = 9 y – 4 3 = 9 y – 4 3 3 · 3 · Multiply to undo division. = 9 y – 4 3 Course 3 10-1 Solving Two-Step Equations
20. Try This : Example 2A A. + 5 = 29 Solve. Think: First the variable is divided by 4, and then 5 is added . To isolate the variable, subtract 5, and then multiply by 4 . – 5 – 5 Subtract to undo addition. n = 96 n 4 + 5 = 29 n 4 Multiply to undo division. 4 = 4 24 n 4 Course 3 10-1 Solving Two-Step Equations
21. Substitute 96 into the original equation. Check Try This : Example 2A Continued + 5 = 29 n 4 ? + 5 = 29 96 4 24 + 5 = 29 ? Course 3 10-1 Solving Two-Step Equations
22. Try This : Example 2B B. 4.8 = –2.3 m + 0.2 Think: First the variable is multiplied by –2.3, and then 0.2 is added . To isolate the variable, subtract 0.2, and then divide by –2.3 . 4.8 = –2.3 m + 0.2 – 0.2 –0.2 Subtract to undo addition. 4.6 = –2.3 m Divide to undo multiplication. – 2 = m – 2.3 –2.3 4.6 = –2.3 m Course 3 10-1 Solving Two-Step Equations
23. Try This : Example 2C Think: First 2 is subtracted from the variable, and then the result is divided by 4 . To isolate the variable, multiply by 4, and then add 2 . y – 2 = 32 + 2 + 2 Add to undo subtraction. y = 34 C. = 8 y – 2 4 = 8 y – 2 4 4 · 4 · Multiply to undo division. = 8 y – 2 4 Course 3 10-1 Solving Two-Step Equations
24. Solve. 1. – 3 = 10 2. 7 y + 25 = –24 3. –8.3 = –3.5 x + 13.4 4. = 3 5. The cost for a new cell phone plan is $39 per month plus a one-time start-up fee of $78. If you are charged $1014, how many months will the contract last? Lesson Quiz y = –7 x = –117 Insert Lesson Title Here x = 6.2 y = 28 24 months x – 9 y + 5 11 Course 3 10-1 Solving Two-Step Equations
25. Warm Up Problem of the Day Lesson Presentation 10-2 Solving Multistep Equations Course 3
26. Warm Up Solve. 1. 3 x = 102 2. = 15 3. z – 100 = –1 4. 1.1 + 5 w = 98.6 x = 34 y = 225 z = 99 w = 19.5 Course 3 10-2 Solving Multistep Equations y 15
27. Problem of the Day Ana has twice as much money as Ben, and Ben has three times as much as Clio. Together they have $160. How much does each person have? Ana, $96; Ben, $48; Clio, $16 Course 3 10-2 Solving Multistep Equations
32. Solve. 9 x + 5 + 4 x – 2 = 42 Try This : Example 1 13 x + 3 = 42 Combine like terms. – 3 – 3 Subtract to undo addition. 13 x = 39 x = 3 Divide to undo multiplication. Course 3 10-2 Solving Multistep Equations 39 13 13 x 13 =
33. Check Try This : Example 1 Continued 9 x + 5 + 4 x – 2 = 42 Substitute 3 for x. 9 (3) + 5 + 4 (3) – 2 = 42 ? 27 + 5 + 12 – 2 = 42 ? 42 = 42 ? Course 3 10-2 Solving Multistep Equations
34. If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) to clear the fractions before you isolate the variable. Course 3 10-2 Solving Multistep Equations
35. Solve. A. + = – Additional Example 2: Solving Equations That Contain Fractions Multiply both sides by 4 to clear fractions, and then solve. ( ) ( ) 5 n + 7 = –3 Distributive Property. 3 4 7 4 5 n 4 7 4 – 3 4 5 n 4 4 + = 4 ( ) ( ) ( ) 5n 4 7 4 – 3 4 4 + 4 = 4 Course 3 10-2 Solving Multistep Equations
36. Additional Example 2 Continued 5 n + 7 = –3 – 7 –7 Subtract to undo addition. 5 n = –10 n = –2 5 n 5 – 10 5 = Divide to undo multiplication. Course 3 10-2 Solving Multistep Equations
37. Insert Lesson Title Here The least common denominator (LCD) is the smallest number that each of the denominators will divide into. Remember! Course 3 10-2 Solving Multistep Equations
38. Solve. B. + – = Additional Example 2B: Solving Equations That Contain Fractions The LCD is 18. 14 x + 9 x – 34 = 12 23 x – 34 = 12 Combine like terms. Distributive Property. Multiply both sides by the LCD. 2 3 x 2 7 x 9 17 9 18 ( ) + 18 ( ) – 18 ( ) = 18 ( ) 7 x 9 x 2 17 9 2 3 ( ) x 2 2 3 7 x 9 17 9 18 + – = 18 Course 3 10-2 Solving Multistep Equations
39. Additional Example 2B Continued 23 x = 46 x = 2 + 34 + 34 Add to undo subtraction. 23 x – 34 = 12 Combine like terms. = 23 x 23 46 23 Divide to undo multiplication. Course 3 10-2 Solving Multistep Equations
41. Solve. A. + = – Try This : Example 2A Multiply both sides by 4 to clear fractions, and then solve. ( ) ( ) 3 n + 5 = –1 Distributive Property. 1 4 5 4 3 n 4 5 4 – 1 4 3 n 4 4 + = 4 ( ) ( ) ( ) 3n 4 5 4 – 1 4 4 + 4 = 4 Course 3 10-2 Solving Multistep Equations
42. Try This : Example 2A Continued 3 n + 5 = –1 – 5 –5 Subtract to undo addition. 3 n = –6 n = –2 3 n 3 – 6 3 = Divide to undo multiplication. Course 3 10-2 Solving Multistep Equations
43. Solve. B. + – = Try This : Example 2B The LCD is 9. 5 x + 3 x – 13 = 3 8 x – 13 = 3 Combine like terms. Distributive Property. Multiply both sides by the LCD. 1 3 x 3 5 x 9 13 9 9 ( ) + 9 ( ) – 9 ( ) = 9 ( ) 5 x 9 x 3 13 9 1 3 ( ) x 3 1 3 5 x 9 13 9 9 + – = 9 ( ) Course 3 10-2 Solving Multistep Equations
44. 8 x = 16 x = 2 + 13 + 13 Add to undo subtraction. 8 x – 13 = 3 Combine like terms. Try This : Example 2B Continued = 8 x 8 16 8 Divide to undo multiplication. Course 3 10-2 Solving Multistep Equations
45. Check + – = Try This : Example 2B Continued 3 9 3 9 = ? x 3 5 x 9 13 9 1 3 1 3 Substitute 2 for x. 5 (2) 9 + – = (2) 3 13 9 ? 1 3 10 9 + – = 2 3 13 9 ? The LCD is 9. 3 9 10 9 + – = 6 9 13 9 ? Course 3 10-2 Solving Multistep Equations
46. When Mr. and Mrs. Harris left for the mall, Mrs. Harris had twice as much money as Mr. Harris had. While shopping, Mrs. Harris spent $54 and Mr. Harris spent $26. When they arrived home, they had a total of $46. How much did Mr. Harris have when he left home? Additional Example 3: Money Application Let h represent the amount of money that Mr. Harris had when he left home. So Mrs. Harris had 2 h when she left home. h + 2 h – 26 – 54 = 46 Mr. Harris $+ Mrs. Harris $ – Mr. Harris spent – Mrs. Harris spent = amount left Course 3 10-2 Solving Multistep Equations
47. Additional Example 3 Continued 3 h – 80 = 46 Combine like terms. 3 h = 126 Divide both sides by 3. h = 42 Mr. Harris had $42 when he left home. + 80 +80 Add 80 to both sides. 3 h 3 126 3 = Course 3 10-2 Solving Multistep Equations
48. When Mr. and Mrs. Wesner left for the store, Mrs. Wesner had three times as much money as Mr. Wesner had. While shopping, Mr. Wesner spent $50 and Mrs. Wesner spent $25. When they arrived home, they had a total of $25. How much did Mr. Wesner have when he left home? Try This : Example 3 Let h represent the amount of money that Mr. Wesner had when he left home. So Mrs. Wesner had 3 h when she left home. h + 3 h – 50 – 25 = 25 Mr. Wesner $ + Mrs. Wesner $ – Mr. Wesner spent – Mrs. Wesner spent = amount left Course 3 10-2 Solving Multistep Equations
49. Try This : Example 3 Continued 4 h – 75 = 25 Combine like terms. + 75 +75 Add 75 to both sides. 4 h = 100 Divide both sides by 4. h = 25 Mr. Wesner had $25 when he left home. 4 h 4 100 4 = Course 3 10-2 Solving Multistep Equations
50. Solve. 1. 6 x + 3 x – x + 9 = 33 2. –9 = 5 x + 21 + 3 x 3. + = 5. Linda is paid double her normal hourly rate for each hour she works over 40 hours in a week. Last week she worked 52 hours and earned $544. What is her hourly rate? Lesson Quiz x = –3.75 x = 3 Insert Lesson Title Here x = 28 4. – = $8.50 5 8 x 8 33 8 6 x 7 2 x 21 25 21 x = 1 9 16 Course 3 10-2 Solving Multistep Equations
51. Warm Up Problem of the Day Lesson Presentation 10-3 Solving Equations with Variables on Both Sides Course 3
52. Warm Up Solve. 1. 2 x + 9 x – 3 x + 8 = 16 2. – 4 = 6 x + 22 – 4 x 3. + = 5 4. – = 3 x = 1 x = -13 x = 34 x = 50 Course 3 10-3 Solving Equations with Variables on Both Sides 2 7 x 7 7 1 9 x 16 2 x 4 1 8
53. Problem of the Day An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle? 22.5 in. Course 3 10-3 Solving Equations with Variables on Both Sides
54. Learn to solve equations with variables on both sides of the equal sign. Course 3 10-3 Solving Equations with Variables on Both Sides
55. Some problems produce equations that have variables on both sides of the equal sign. Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation. Course 3 10-3 Solving Equations with Variables on Both Sides
56. Solve. A. 4 x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4 x + 6 = x – 4 x – 4 x 6 = –3 x Subtract 4x from both sides. Divide both sides by – 3. – 2 = x 6 – 3 – 3 x – 3 = Course 3 10-3 Solving Equations with Variables on Both Sides
57. Solve. B. 9 b – 6 = 5 b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9 b – 6 = 5 b + 18 – 5 b – 5 b 4 b – 6 = 18 Subtract 5b from both sides. Divide both sides by 4. b = 6 + 6 + 6 4 b = 24 Add 6 to both sides. 4 b 4 24 4 = Course 3 10-3 Solving Equations with Variables on Both Sides
58. Solve. C. 9 w + 3 = 5 w + 7 + 4 w Additional Example 1C: Solving Equations with Variables on Both Sides 9 w + 3 = 5 w + 7 + 4 w 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true. 9 w + 3 = 9 w + 7 Combine like terms. – 9 w – 9 w Subtract 9w from both sides. Course 3 10-3 Solving Equations with Variables on Both Sides
59. Solve. A. 5 x + 8 = x Try This : Example 1A 5 x + 8 = x – 5 x – 5 x 8 = –4 x Subtract 4x from both sides. Divide both sides by – 4. – 2 = x 8 – 4 – 4 x – 4 = Course 3 10-3 Solving Equations with Variables on Both Sides
60. Solve. B. 3 b – 2 = 2 b + 12 3 b – 2 = 2 b + 12 – 2 b – 2 b b – 2 = 12 Subtract 2b from both sides. + 2 + 2 b = 14 Add 2 to both sides. Try This : Example 1B Course 3 10-3 Solving Equations with Variables on Both Sides
61. Solve. C. 3 w + 1 = 10 w + 8 – 7 w 3 w + 1 = 10 w + 8 – 7 w 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true. Try This : Example 1C 3 w + 1 = 3 w + 8 Combine like terms. – 3 w – 3 w Subtract 3w from both sides. Course 3 10-3 Solving Equations with Variables on Both Sides
62. To solve multistep equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable. Course 3 10-3 Solving Equations with Variables on Both Sides
63. Solve. A. 10 z – 15 – 4 z = 8 – 2 z - 15 Additional Example 2A: Solving Multistep Equations with Variables on Both Sides 10 z – 15 – 4 z = 8 – 2 z – 15 + 15 +15 6 z – 15 = –2 z – 7 Combine like terms. + 2 z + 2 z Add 2z to both sides. 8 z – 15 = – 7 8 z = 8 z = 1 Add 15 to both sides. Divide both sides by 8. 8 z 8 8 8 = Course 3 10-3 Solving Equations with Variables on Both Sides
64. B. Additional Example 2B: Solving Multistep Equations with Variables on Both Sides Multiply by the LCD. 4 y + 12 y – 15 = 20 y – 14 16 y – 15 = 20 y – 14 Combine like terms. + – = y – y 5 3 4 3 y 5 7 10 y 5 3 4 3 y 5 7 10 + – = y – 20 ( ) = 20 ( ) y 5 3 4 3 y 5 7 10 + – y – 20 ( ) + 20 ( ) – 20 ( ) = 20( y ) – 20 ( ) y 5 3 y 5 3 4 7 10 Course 3 10-3 Solving Equations with Variables on Both Sides
65. Additional Example 2B Continued Add 14 to both sides. – 15 = 4 y – 14 – 1 = 4 y + 14 + 14 Divide both sides by 4. 16 y – 15 = 20 y – 14 – 16 y – 16 y Subtract 16y from both sides. – 1 4 4 y 4 = -1 4 = y Course 3 10-3 Solving Equations with Variables on Both Sides
66. Solve. A. 12 z – 12 – 4 z = 6 – 2 z + 32 Try This : Example 2A 12 z – 12 – 4 z = 6 – 2 z + 32 + 12 +12 8 z – 12 = –2 z + 38 Combine like terms. + 2 z + 2 z Add 2z to both sides. 10 z – 12 = + 38 10 z = 50 z = 5 Add 12 to both sides. Divide both sides by 10. 10 z 50 10 10 = Course 3 10-3 Solving Equations with Variables on Both Sides
67. B. Multiply by the LCD. 6 y + 20 y + 18 = 24 y – 18 26 y + 18 = 24 y – 18 Combine like terms. + + = y – Try This : Example 2B y 4 3 4 5 y 6 6 8 y 4 3 4 5 y 6 6 8 + + = y – 24 ( ) = 24 ( ) y 4 3 4 5 y 6 6 8 + + y – 24 ( ) + 24 ( ) + 24 ( ) = 24( y ) – 24 ( ) y 4 5 y 6 3 4 6 8 Course 3 10-3 Solving Equations with Variables on Both Sides
68. Subtract 18 from both sides. 2 y + 18 = – 18 2 y = –36 – 18 – 18 Divide both sides by 2. y = –18 26 y + 18 = 24 y – 18 – 24 y – 24 y Subtract 24y from both sides. Try This : Example 2B Continued – 36 2 2 y 2 = Course 3 10-3 Solving Equations with Variables on Both Sides
69. Additional Example 3: Consumer Application Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday? Course 3 10-3 Solving Equations with Variables on Both Sides
70. Additional Example 3 Continued First solve for the price of one doughnut. 1.25 + 2 d = 0.50 + 5 d Let d represent the price of one doughnut. – 2 d – 2 d 1.25 = 0.50 + 3 d Subtract 2d from both sides. – 0.50 – 0.50 Subtract 0.50 from both sides. 0.75 = 3 d Divide both sides by 3. 0.25 = d The price of one doughnut is $0.25. 0.75 3 3 d 3 = Course 3 10-3 Solving Equations with Variables on Both Sides
71. Additional Example 3 Continued Now find the amount of money Jamie spends each morning. 1.25 + 2 d Choose one of the original expressions. Jamie spends $1.75 each morning. 1.25 + 2 (0.25) = 1.75 Let n represent the number of doughnuts. Find the number of doughnuts Jamie buys on Tuesday. 0.25 n = 1.75 n = 7; Jamie bought 7 doughnuts on Tuesday. Divide both sides by 0.25. 0.25 n 0.25 1.75 0.25 = Course 3 10-3 Solving Equations with Variables on Both Sides
72. Try This : Example 3 Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays? Course 3 10-3 Solving Equations with Variables on Both Sides
73. Try This : Example 3 Continued First solve for distance around the track. 2 x + 4 = 4 x + 2 Let x represent the distance around the track. – 2 x – 2 x 4 = 2 x + 2 Subtract 2x from both sides. – 2 – 2 Subtract 2 from both sides. 2 = 2 x Divide both sides by 2. 1 = x The track is 1 mile around. 2 2 2 x 2 = Course 3 10-3 Solving Equations with Variables on Both Sides
74. Try This : Example 3 Continued Now find the total distance Helene walks each day. 2 x + 4 Choose one of the original expressions. Helene walks 6 miles each day. 2 (1) + 4 = 6 Let n represent the number of 1-mile laps. Find the number of laps Helene walks on Saturdays. 1 n = 6 Helene walks 6 laps on Saturdays. n = 6 Course 3 10-3 Solving Equations with Variables on Both Sides
75. Lesson Quiz Solve. 1. 4 x + 16 = 2 x 2. 8 x – 3 = 15 + 5 x 3. 2(3 x + 11) = 6 x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = 6 x = –8 Insert Lesson Title Here no solution x = 36 An orange has 45 calories. An apple has 75 calories. 1 4 1 2 Course 3 10-3 Solving Equations with Variables on Both Sides
76. Warm Up Problem of the Day Lesson Presentation 10-4 Solving Multistep Inequalities Course 3
77. Warm Up Solve. 1. 6 x + 36 = 2 x 2. 4 x – 13 = 15 + 5 x 3. 5( x – 3) = 2 x + 3 4. + x = x = –9 x = –28 x = 6 Course 3 10-4 Solving Multistep Inequalities 7 8 3 16 11 16 x = –
78. Problem of the Day Find an integer x that makes the following two inequalities true: 4 < x 2 < 16 and x < 2.5 x = –3 Course 3 10-4 Solving Multistep Inequalities
79. Learn to solve two-step inequalities and graph the solutions of an inequality on a number line . Course 3 10-4 Solving Multistep Inequalities
80. Solving a multistep inequality uses the same inverse operations as solving a multistep equation. Multiplying or dividing the inequality by a negative number reverses the inequality symbol. Course 3 10-4 Solving Multistep Inequalities
81. Solve and graph. Additional Example 1A: Solving Multistep Inequalities A. 4 x + 1 > 13 4 x + 1 > 13 – 1 – 1 Subtract 1 from both sides. 4 x > 12 Divide both sides by 4. x > 3 4 x 4 > 12 4 1 2 3 4 5 6 7 Course 3 10-4 Solving Multistep Inequalities
82. Additional Example 1B: Solving Multistep Inequalities B. – 7 < 3 x + 8 – 7 < 3 x + 8 – 8 – 8 Subtract 8 from both sides. – 15 < 3 x Divide both sides by 3. – 5 < x – 15 3 < 3 x 3 -7 -6 -5 -4 -3 -2 -1 Course 3 10-4 Solving Multistep Inequalities
83. Additional Example 1C: Solving Multistep Inequalities C. -9 x + 7 25 – 9 x + 7 25 – 7 – 7 Subtract 7 from both sides. – 9 x 18 Divide each side by –9; change to . x – 2 – 9 x – 9 18 – 9 -6 -5 -4 -3 -2 -1 0 Course 3 10-4 Solving Multistep Inequalities
84. Solve and graph. Try This : Example 1A A. 5 x + 2 > 12 5 x + 2 > 12 – 2 – 2 Subtract 2 from both sides. 5 x > 10 Divide both sides by 5. x > 2 5 x 5 > 10 5 1 2 3 4 5 6 7 Course 3 10-4 Solving Multistep Inequalities
85. B. – 5 < 2 x + 9 – 5 < 2 x + 9 – 9 – 9 Subtract 9 from both sides. – 14 < 2 x Divide both sides by 2. – 7 < x Try This : Example 1B – 14 2 < 2 x 2 -7 -6 -5 -4 -3 -2 -1 Course 3 10-4 Solving Multistep Inequalities
86. C. -4 x + 2 18 – 4 x + 2 18 – 2 – 2 Subtract 2 from both sides. – 4 x 16 Divide each side by –4; change to . x – 4 Try This : Example 1C – 4 x – 4 16 – 4 -6 -5 -4 -3 -2 -1 0 Course 3 10-4 Solving Multistep Inequalities
87. Solve and graph. Additional Example 2A: Solving Multistep Inequalities A. 10 x + 21 – 4 x < –15 10 x + 21 – 4 x < –15 – 21 – 21 Subtract 21 from both sides. Divide both sides by 6. x < –6 6 x + 21 < –15 Combine like terms. 6 x < –36 6 x 6 < – 36 6 -8 -7 -6 -5 -4 -3 -2 Course 3 10-4 Solving Multistep Inequalities
88. Additional Example 2B: Solving Multistep Inequalities Multiply by LCD, 20. 8 x + 15 18 – 15 – 15 Subtract 15 from both sides. 8 x 3 B. + 2x 5 3 4 9 10 + 2 x 5 3 4 9 10 20 ( + ) 20 ( ) 2 x 5 3 4 9 10 20 ( ) + 20 ( ) 20 ( ) 2 x 5 3 4 9 10 Course 3 10-4 Solving Multistep Inequalities
89. Additional Example 2 Continued 8 x 3 x 3 8 8 x 8 3 8 Divide both sides by 8. Course 3 10-4 Solving Multistep Inequalities 0 1 3 8
90. Additional Example 2C: Solving Multistep Inequalities C. 8 x + 8 > 11 x – 1 8 x + 8 > 11 x – 1 – 8 x – 8 x Subtract 8x from both sides. 8 > 3 x – 1 Add 1 to each side. 3 > x +1 +1 9 > 3 x Divide both sides by 3. 9 3 > 3 x 3 -1 0 1 2 3 4 5 Course 3 10-4 Solving Multistep Inequalities
91. Solve and graph. Try This : Example 2A A. 15 x + 30 – 5 x < –10 15 x + 30 – 5 x < –10 – 30 – 30 Subtract 30 from both sides. Divide both sides by 10. x < –4 10 x + 30 < – 10 Combine like terms. 10 x < –40 10 x 10 < – 40 10 -8 -7 -6 -5 -4 -3 -2 Course 3 10-4 Solving Multistep Inequalities
92. Try This : Example 2B Multiply by LCD, 20. 12 x + 5 10 – 5 – 5 Subtract 5 from both sides. 12 x 5 B. + 3x 5 1 4 5 10 + 3 x 5 1 4 5 10 20 ( + ) 20 ( ) 3 x 5 1 4 5 10 20 ( ) + 20 ( ) 20 ( ) 3 x 5 1 4 5 10 Course 3 10-4 Solving Multistep Inequalities
93. Try This : Example 2B Continued 12 x 5 x 5 12 12 x 12 5 12 Divide both sides by 12. 0 5 12 Course 3 10-4 Solving Multistep Inequalities
94. Try This : Example 2C C. 4 x + 3 > 8 x – 1 4 x + 3 > 8 x – 1 – 4 x – 4 x Subtract 4x from both sides. 3 > 4 x – 1 Add 1 to each side. 1 > x +1 +1 4 > 4 x Divide both sides by 4. 4 4 > 4 x 4 -1 0 1 2 3 4 5 Course 3 10-4 Solving Multistep Inequalities
95. Additional Example 3: Business Application A school’s Spanish club is selling bumper stickers. They bought 100 bumper stickers for $55, and they have to give the company 15 cents for every sticker sold. If they plan to sell each bumper sticker for $1.25, how many do they have to sell to make a profit? Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost. R > C Course 3 10-4 Solving Multistep Inequalities
96. Additional Example 3 Continued The revenue from selling x bumper stickers at $1.25 each is 1.25 x . The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 55 + 0.15 x . Substitute the expressions for R and C. 1.25 x > 55 + 0.15 x Let x represent the number of bumper stickers sold. Fixed cost is $55. Unit cost is 15 cents. Course 3 10-4 Solving Multistep Inequalities
97. Additional Example 3 Continued – 0.15x – 0.15 x Subtract 0.15x from both sides. 1.10 x > 55 x > 50 The Spanish club must sell more than 50 bumper stickers to make a profit. Divide both sides by 1.10. 1.25 x > 55 + 0.15 x 1.10 x 1.10 55 1.10 > Course 3 10-4 Solving Multistep Inequalities
98. Try This : Example 3 R > C A school’s Spanish club is selling bumper stickers. They bought 200 bumper stickers for $45, and they have to give the company 25 cents for every sticker sold. If they plan to sell each bumper sticker for $2.50, how many do they have to sell to make a profit? Let R represent the revenue and C represent the cost. In order for the Spanish club to make a profit, the revenue must be greater than the cost. Course 3 10-4 Solving Multistep Inequalities
99. Try This : Example 3 Continued The revenue from selling x bumper stickers at $2.50 each is 2.5 x . The cost of selling x bumper stickers is the fixed cost plus the unit cost times the number of bumper stickers sold, or 45 + 0.25 x . Substitute the expressions for R and C. 2.5 x > 45 + 0.25 x Let x represent the number of bumper stickers sold. Fixed cost is $45. Unit cost is 25 cents. Course 3 10-4 Solving Multistep Inequalities
100. – 0.25x – 0.25 x Subtract 0.25x from both sides. 2.25 x > 45 x > 20 The Spanish club must sell more than 20 bumper stickers to make a profit. Divide both sides by 2.25. 2.5 x > 45 + 0.25 x Try This : Example 3 Continued 2.25 x 2.25 45 2.25 > Course 3 10-4 Solving Multistep Inequalities
101. Lesson Quiz: Part 1 Solve and graph. 1. 4 x – 6 > 10 2. 7 x + 9 < 3 x – 15 3. w – 3 w < 32 4. w + x < –6 x > 4 Insert Lesson Title Here w > –16 2 3 1 4 1 2 w 3 8 1 2 3 4 5 6 7 -10 -9 -8 -7 -6 -5 -4 -18 -17 -16 -15 -14 -13 -12 0 3 8 Course 3 10-4 Solving Multistep Inequalities
102. Lesson Quiz: Part 2 5. Antonio has budgeted an average of $45 a month for entertainment. For the first five months of the year he has spent $48, $39, $60, $48, and $33. How much can Antonio spend in the sixth month without exceeding his average budget? no more than $42 Course 3 10-4 Solving Multistep Inequalities
103. Warm Up Problem of the Day Lesson Presentation 10-5 Solving for a Variable Course 3
104. Warm Up Solve. 1. 8 x – 9 = 23 2. 9 x + 12 = 4 x + 37 3. 6 x – 8 = 7 x + 3 4. x + 3 = x = 4 x = 5 x = –11 Course 3 10-5 Solving for a Variable 1 2 1 8 23 4 x = – , or –5 3 4
105. Problem of the Day The formula A = 4 r 2 gives the surface area of a geometric figure. Solve the formula for r . Can you identify what the geometric figure is? sphere Course 3 10-5 Solving for a Variable
106. Learn to solve an equation for a variable . Course 3 10-5 Solving for a Variable
107. If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation. Course 3 10-5 Solving for a Variable
108. Solve for the indicated variable. Additional Example 1A: Solving for a Variable by Addition or Subtraction A. Solve a – b + 1 = c for a . a – b + 1 = c + b – 1 + b – 1 Add b and subtract 1 from both sides. a = c + b – 1 Isolate a. Course 3 10-5 Solving for a Variable
109. Solve for the indicated variable. Additional Example 1B: Solving for a Variable by Addition or Subtraction B. Solve a – b + 1 = c for b . a – b + 1 = c – a – 1 – a – 1 Subtract a and 1 from both sides. – b = c – a – 1 Isolate b. Multiply both sides by –1. – 1 (– b ) = –1 ( c – a – 1) b = – c + a + 1 Isolate b. Course 3 10-5 Solving for a Variable
110. Solve for the indicated variable. Try This : Example 1A A. Solve y – b + 3 = c for y . y – b + 3 = c + b – 3 + b – 3 Add b and subtract 3 from both sides. y = c + b – 3 Isolate y. Course 3 10-5 Solving for a Variable
111. Solve for the indicated variable. Try This : Example 1B B. Solve p – w + 4 = f for w . p – w + 4 = f – p – 4 – p – 4 Subtract p and 4 from both sides. – w = f – p – 4 Isolate w. Multiply both sides by –1. – 1 (– w ) = –1 ( f – p – 4) w = – f + p + 4 Isolate w. Course 3 10-5 Solving for a Variable
112. To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values. Course 3 10-5 Solving for a Variable
113. Solve for the indicated variable. Assume all values are positive. Additional Example 2A: Solving for a Variable by Division or Square Roots A. Solve A = s 2 for s . A = s 2 Course 3 10-5 Solving for a Variable √ A = s Isolate s. √ A = √ s 2 Take the square root of both sides.
114. Additional Example 2B: Solving for a Variable by Division or Square Roots B. Solve V = IR for R . V = IR Solve for the indicated variable. Assume all values are positive. = Divide both sides by I. IR I V I I V = R Isolate R. Course 3 10-5 Solving for a Variable
115. Additional Example 2C: Solving for a Variable by Division or Square Roots C. Solve the formula for the area of a trapezoid for h . Assume all values are positive. Multiply both sides by 2. 2 A = h ( b 1 + b 2 ) A = h ( b 1 + b 2 ) 1 2 Write the formula. 2 • A = 2 • h ( b 1 + b 2 ) 1 2 Course 3 10-5 Solving for a Variable 2 A ( b 1 + b 2 ) = h Isolate h. 2 A ( b 1 + b 2 ) h ( b 1 + b 2 ) ( b 1 + b 2 ) = Divide both sides by (b 1 + b 2 ).
116. Solve for the indicated variable. Assume all values are positive. Try This : Example 2A A. Solve w = g 2 + 4 for g . w = g 2 + 4 w – 4 = g 2 – 4 – 4 Subtract 4 from both sides. Take the square root of both sides. Course 3 10-5 Solving for a Variable √ w – 4 = g Isolate g. √ w – 4 = √ g 2
117. B. Solve A = lw for w . A = lw Solve for the indicated variable. Assume all values are positive. Try This : Example 2B = Divide both sides by l. lw l A l l A = w Isolate w. Course 3 10-5 Solving for a Variable
118. C. Solve s = 180( n – 2) for n . s = 180( n – 2) Solve for the indicated variable. Assume all values are positive. Try This : Example 2C Divide both sides by 180. Add 2 to both sides. + 2 + 2 Course 3 10-5 Solving for a Variable s 180( n – 2) 180 180 = s 180 = ( n – 2) s 180 + 2 = n
119. Course 3 10-5 Solving for a Variable To find solutions ( x , y ), choose values for x substitute to find y . Remember!
120. Additional Example 3: Solving for y and Graphing Solve for y and graph 3 x + 2 y = 8. 3 x + 2 y = 8 – 3 x –3 x 2 y = –3 x + 8 x y – 2 7 0 4 2 1 4 – 2 – 3 x + 8 2 2 y 2 = y = + 4 – 3 x 2 Course 3 10-5 Solving for a Variable
121. Additional Example 3 Continued 3 x + 2 y = 8 Course 3 10-5 Solving for a Variable
122. Try This : Example 3 Solve for y and graph 4 x + 3 y = 12. 4 x + 3 y = 12 – 4 x – 4 x 3 y = –4 x + 12 x y – 3 8 0 4 3 0 6 – 4 – 4 x + 12 3 3 y 3 = y = + 4 – 4 x 3 Course 3 10-5 Solving for a Variable
123. Try This : Example 3 Continued x y – 4 –2 2 4 6 10 8 6 2 – 2 – 4 4 4 x + 3 y = 12 – 6 Course 3 10-5 Solving for a Variable
124. Lesson Quiz: Part 1 Solve for the indicated variable. 1. P = R – C for C . 2. P = 2 l + 2 w for l . 3. V = Ah for h . 4. R = for S . C = R - P Insert Lesson Title Here C – Rt = S 1 3 C – S t = h 3 V A = l P – 2 w 2 Course 3 10-5 Solving for a Variable
125. Lesson Quiz: Part 2 5. Solve for y and graph 2 x + 7 y = 14. Insert Lesson Title Here y = – + 2 2 x 7 Course 3 10-5 Solving for a Variable
126. Warm Up Problem of the Day Lesson Presentation 10-6 Systems of Equations Course 3
127. Warm Up Solve for the indicated variable. 1. P = R – C for R 2. V = Ah for A 3. R = for C R = P + C Rt + S = C Course 3 10-6 Systems of Equations 1 3 C – S t = h 3 V A
128. Problem of the Day At an audio store, stereos have 2 speakers and home-theater systems have 5 speakers. There are 30 sound systems with a total of 99 speakers. How many systems are stereo systems and how many are home-theater systems? 17 stereo systems, 13 home-theater systems Course 3 10-6 Systems of Equations
129. Learn to solve systems of equations . Course 3 10-6 Systems of Equations
130. Vocabulary system of equations solution of a system of equations Insert Lesson Title Here Course 3 10-6 Systems of Equations
131. A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs. Course 3 10-6 Systems of Equations
132. Determine if the ordered pair is a solution of the system of equations below. 5 x + y = 7 x – 3 y = 11 Additional Example 1A: Identifying Solutions of a System of Equations A. (1, 2) 5 x + y = 7 7 = 7 x – 3 y = 11 Substitute for x and y. – 5 11 The ordered pair (1, 2) is not a solution of the system of equations. 5 (1) + 2 = 7 ? 1 – 3 (2) = 11 ? Course 3 10-6 Systems of Equations
133. Additional Example 1B: Identifying Solutions of a System of Equations B. (2, –3) 7 = 7 Substitute for x and y. 11 = 11 The ordered pair (2, –3) is a solution of the system of equations. Determine if the ordered pair is a solution of the system of equations below. 5 x + y = 7 x – 3 y = 11 5 x + y = 7 x – 3 y = 11 5 (2) + –3 = 7 ? 2 – 3 (–3) = 11 ? Course 3 10-6 Systems of Equations
134. Additional Example 1C: Identifying Solutions of a System of Equations C. (20, 3) 103 7 Substitute for x and y. 11 = 11 The ordered pair (20, 3) is not a solution of the system of equations. Determine if the ordered pair is a solution of the system of equations below. 5 x + y = 7 x – 3 y = 11 5 x + y = 7 x – 3 y = 11 5 (20) + (3) = 7 ? 20 – 3 (3) = 11 ? Course 3 10-6 Systems of Equations
135. Determine if each ordered pair is a solution of the system of equations below. 4 x + y = 8 x – 4 y = 12 Try This : Example 1A A. (1, 2) 4 x + y = 8 6 8 x – 4 y = 12 Substitute for x and y. – 7 12 The ordered pair (1, 2) is not a solution of the system of equations. 4 (1) + 2 = 8 ? 1 – 4 (2) = 12 ? Course 3 10-6 Systems of Equations
136. Try This : Example 1B Determine if each ordered pair is a solution of the system of equations below. 4 x + y = 8 x – 4 y = 12 B. (2, –3) 5 8 Substitute for x and y. 14 12 The ordered pair (2, –3) is not a solution of the system of equations. 4 x + y = 8 x – 4 y = 12 4 (2) + –3 = 8 ? 2 – 4 (–3) = 12 ? Course 3 10-6 Systems of Equations
137. Try This : Example 1C C. (1, 4) The ordered pair (1, 4) is not a solution of the system of equations. Determine if each ordered pair is a solution of the system of equations below. 4 x + y = 8 x – 4 y = 12 8 = 8 Substitute for x and y. – 15 12 4 x + y = 8 x – 4 y = 12 4 (1) + 4 = 8 ? 1 – 4 (4) = 12 ? Course 3 10-6 Systems of Equations
138. Course 3 10-6 Systems of Equations When solving systems of equations, remember to find values for all of the variables. Helpful Hint
139. Additional Example 2: Solving Systems of Equations Solve the system of equations. y = x – 4 y = 2 x – 9 Solve the equation to find x . x – 4 = 2 x – 9 – x – x Subtract x from both sides. – 4 = x – 9 5 = x + 9 + 9 Add 9 to both sides. y = x – 4 y = 2 x – 9 y = y x – 4 = 2 x – 9 Course 3 10-6 Systems of Equations
140. Additional Example 2 Continued To find y , substitute 5 for x in one of the original equations. y = x – 4 = 5 – 4 = 1 The solution is (5, 1). Check: Substitute 5 for x and 1 for y in each equation. y = x – 4 y = 2 x – 9 1 = 1 1 = 1 1 = 5 – 4 ? 1 = 2 (5) – 9 ? Course 3 10-6 Systems of Equations
141. Try This : Example 2 Solve the system of equations. y = x – 5 y = 2 x – 8 Solve the equation to find x . x – 5 = 2 x – 8 – x – x Subtract x from both sides. – 5 = x – 8 3 = x + 8 + 8 Add 8 to both sides. y = x – 5 y = 2 x – 8 y = y x – 5 = 2 x – 8 Course 3 10-6 Systems of Equations
142. Try This : Example 2 Continued To find y , substitute 3 for x in one of the original equations. y = x – 5 = 3 – 5 = –2 The solution is (3, –2). Check: Substitute 3 for x and –2 for y in each equation. y = x – 5 y = 2 x – 8 – 2 = –2 – 2 = –2 – 2 = 3 – 5 ? – 2 = 2 (3) – 8 ? Course 3 10-6 Systems of Equations
143. To solve a general system of two equations with two variables, you can solve both equations for x or both for y . Course 3 10-6 Systems of Equations
144. Additional Example 3A: Solving Systems of Equations Solve the system of equations. A. x + 2 y = 8 x – 3 y = 13 x + 2 y = 8 x – 3 y = 13 – 2 y –2 y + 3 y + 3 y Solve both equations for x. x = 8 – 2 y x = 13 + 3 y 8 – 2 y = 13 + 3 y + 2 y + 2 y 8 = 13 + 5 y Add 2y to both sides. Course 3 10-6 Systems of Equations
145. Additional Example 3A Continued 8 = 13 + 5 y – 13 –13 – 5 = 5 y Subtract 13 from both sides. Divide both sides by 5. – 1 = y x = 8 – 2 y = 8 – 2 (–1) Substitute – 1 for y. = 8 + 2 = 10 The solution is (10, –1). – 5 5 5 y 5 = Course 3 10-6 Systems of Equations
146. Course 3 10-6 Systems of Equations You can choose either variable to solve for. It is usually easiest to solve for a variable that has a coefficient of 1. Helpful Hint
147. Additional Example 3B: Solving Systems of Equations Solve the system of equations. B. 3 x – 3 y = -3 2 x + y = -5 3 x – 3 y = –3 2 x + y = –5 – 3 x –3 x –2 x –2 x Solve both equations for y. – 3 y = –3 – 3 x y = –5 – 2 x y = 1 + x 1 + x = –5 – 2 x – 3 – 3 3 x – 3 – 3 y – 3 = – Course 3 10-6 Systems of Equations
148. Additional Example 3B Continued + 2 x + 2 x Add 2x to both sides. 1 + 3 x = –5 – 1 –1 3 x = –6 1 + x = –5 – 2 x Subtract 1 from both sides. Divide both sides by 3. x = –2 y = 1 + x = 1 + –2 = –1 Substitute –2 for x. The solution is (–2, –1). – 6 3 3 x 3 = Course 3 10-6 Systems of Equations
149. Try This : Example 3A Solve the system of equations. A. x + y = 5 3 x + y = –1 x + y = 5 3 x + y = –1 – x – x – 3 x – 3 x Solve both equations for y. y = 5 – x y = –1 – 3 x 5 – x = –1 – 3 x + x + x 5 = –1 – 2 x Add x to both sides. Course 3 10-6 Systems of Equations
150. Try This : Example 3A Continued 5 = –1 – 2 x + 1 + 1 6 = –2 x Add 1 to both sides. Divide both sides by –2. – 3 = x y = 5 – x = 5 – (–3) Substitute – 3 for x. = 5 + 3 = 8 The solution is (–3, 8). Course 3 10-6 Systems of Equations
151. Try This : Example 3B Solve the system of equations. B. x + y = –2 –3 x + y = 2 x + y = –2 –3 x + y = 2 – x – x + 3 x + 3 x Solve both equations for y. y = –2 – x y = 2 + 3 x – 2 – x = 2 + 3 x Course 3 10-6 Systems of Equations
152. + x + x Add x to both sides. – 2 = 2 + 4 x – 2 –2 – 4 = 4 x – 2 – x = 2 + 3 x Subtract 2 from both sides. Divide both sides by 4. – 1 = x y = 2 + 3 x = 2 + 3 (–1) = –1 Substitute –1 for x. The solution is (–1, –1). Try This : Example 3B Continued Course 3 10-6 Systems of Equations
153. Lesson Quiz 1. Determine if the ordered pair (2, 4) is a solution of the system. y = 2 x ; y = –4 x + 12 Solve each system of equations. 2. y = 2 x + 1; y = 4 x 3. 6 x – y = –15; 2 x + 3 y = 5 4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers. yes Insert Lesson Title Here ( – 2,3) 15 and 8 ( , 2 ) 1 2 Course 3 10-6 Systems of Equations
