1. Chapter 2 - Equations Algebra I

2. Table of Contents 2.1- Solving Equations by Adding or Subtracting 2.1 2.2- Solving Equations by Multiplying and Dividing 2.2 2.3- Solving Two-Step and Multi-Step Equations 2.3 2.4- Solving Equations with Variables on Both Sides 2.4 2.5- Solving for a Variable 2.5 2.6- Solving Absolute-Value Equations 2.6 2.7- Rates, Ratios, and Proportions 2.7 2.9- Percents 2.9 2.10- Applications of Percents 2.10

3. 2.1 - Solving Equations by Adding or Subtracting Algebra I

4. An equation is a mathematical statement that two expressions are equal. A solution of an equation is a value of the variable that makes the equation true. 2-1 Algebra 1 (bell work) To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.

5. Solve the equation. Check your answer. y – 8 = 24 + 8 + 8 y = 32 Check y – 8 = 24 32 – 8 24 24 2-1 = z 3 4 + 7 16 + 7 = z – 7 16 5 Check= z – 7 16 5 3 4 5 7 – 5 5 Example 1Solving Equations by Using Addition

6. –6 = k – 6 + 6 + 6 0 = k Check –6 = k – 6 –6 0 – 6 –6 2-1 p = 3 11 + 5 + 5 Solve –+ p = – 2 11 5 Check + p = – 2 11 5 – 2 5 – – 3 + 2 – 2 –

7. Solve the equation. Check your answer. m + 17 = 33 – 17 –17 m = 16 Check m + 17 = 33 16 + 17 33 33 2-1Example 2Solving Equations using Subtraction d + = 1 – 1 2 – 1 2 d = 1 2 1 2

8. Math Joke Parent: Why do you have that sheet of paper in a bowl of water? Student: It’s my homework, I’m trying to dissolve an equation

9. Solve –2.3 + m = 7. –2.3 + m = 7 +2.3 + 2.3 m = 9.3 Check –2.3 + m = 7 –2.3 + 9.3 7 7 2-1Example 3Solving Equations by Adding the Opposite z = 2 + 3 4 + 3 4 Solve Check + z = 5 4 3 4 – – 5 3 4 4 + 2 5 4 5 4 – + z = 5 4 3 4

10. Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population. + 275 p =1125 p – d = c original population minus current population decrease in population is p – 275 = 850 p – d = c The original population was 1125 people. 2-1Example 4Application

11. A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute. 2-1

12. a + r = 220 age added to 220maximum heart rate is – 185 a = 35 a + 185 = 220 a + r = 220 A person whose maximum heart rate is 185 beats per minute would be 35 years old. 2-1

13. HW pg. 80 2.1 – 21-35 (Odd), 46, 47, 49-63 (Odd), 66, 80, 83, 86 – Must show check step 2.1

14. 2.2 - Solving Equations by Multiplying or Dividing Algebra I

15. Solve the equation. –24 = j –8 –8 = j 3 –8 –24 3 Check –8 = j 3 p = 50 10 = 10 p 5 10 50 5 Check = 10 p 5 2-2Example 1Solving Equations by Using Multiplication

16. y = 12 108 9y = 108 9(12) 108 Check 9y = 108 4 = c 16 16 = 4c 16 4(4) Check 16 = 4c 2-2Example 2Solving Equations by Using Division

17. Math Joke Teacher: Why don’t you have your homework today? Student: I divided by zero and the paper vanished into thin air! 2-2

18. Solve the equation. w =  24  20 w =  20 5 6 Check w =  20 5 6  20 = z 3 2 Check 1 8 3 16 = z 3 16 3 3 1 8 2-2Example 3Solving Equations That Contain Fractions

19. Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year. 1 4 m = $1140 Ciro earned $1140 mowing lawns. 2-2Example 4Application

20. The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began. 15 = h The plane was flying at 15,000 ft when the descent began. 2-2

21. HW pg. 87 2.2- – 21-33 (Odd), 19, 20, 45, 46, 49, 52-55, 76 – Ch: 56, 58, 60, 65 – Must show check step

22. 2.3 - Solving Two-Step and Multi- Step Equations Algebra I

23. 2-3 Cost per CD Total cost Cost of discount card Notice that this equation contains multiplication and addition. Equations that contain more than one operation require more than one step to solve. Identify the operations in the equation and the order in which they are applied to the variable. Then use inverse operations and work backward to undo them one at a time. Algebra 1 (bell work)Just Read Operations in the Equation 1. 1. First c is multiplied by 3.95. 2. 2. Then 19.95 is added. To Solve 1. 1. Subtract 19.95 from both sides of the equation. 2. 2. Then divide both sides by 3.95.

24. Solve 18 = 4a + 10. 18 = 4a + 10 –10 – 10 8 = 4a 44 2 = a Solve 5t – 2 = –32. 5t – 2 = –32 + 2 5t = –30 55 t = –6 2-3Example 1Solving Two-Step Equations

25. –2 n = 0 Solve 2-3Example 2Solving Two-Step Equations that Contain Fractions

26. Solve. 3y – 18 = 14 +18 3y = 32 3 3y = 32 2-3

27. 8r + 9 = 7 –9 8r = –2 8 2-3

28. Math Joke Q: How do equation get into shape? A: They do multi-step aerobics 2-3

29. Solve 8x – 21 + 5x = –15. 8x – 21 – 5x = –15 8x – 5x – 21 = –15 3x – 21 = –15 + 21 +21 3x = 6 x = 2 2-3Example 3Simplifying Before Solving Equations Solve 10y – (4y + 8) = –20 10y + (–1)(4y + 8) = –20 10y + (–1)(4y) + (–1)( 8) = –20 10y – 4y – 8 = –20 + 8 y = –2

30. If 4a + 0.2 = 5, find the value of a – 1. Step 1 Find the value of a. 4a + 0.2 = 5 –0.2 –0.2 4a = 4.8 a = 1.2 Step 2 Find the value of a – 1. 1.2 – 1 0.2 2-3 If 3d – (9 – 2d) = 51, find the value of 3d. Step 1 Find the value of d. 3d – (9 – 2d) = 51 +9 +9 5d = 60 d = 12 3d – 9 + 2d = 51 5d – 9 = 51 d = 12 Step 2 Find the value of 3d. 3(12)=36 Example 5Solving Equations to find an Indicated Value

31. HW pg. 96 2.3- – 25-41 (Odd), 19, 42-47, 68, 81-87 (Odd) – Ch: 50, 52, 53, 67, 80

32. 2.4 - Solving Equations with Variables on Both Sides Algebra I

33. Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 –5n 2n – 2 = 6 2n = 8 + 2 n = 4 Solve 4b + 2 = 3b. 4b + 2 = 3b –3b b + 2 = 0 b = –2 – 2 – 2 2-4Example 1Solving Equations with Variables on Both Sides

34. Solve 0.5 + 0.3y = 0.7y – 0.3. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y 2-4

35. Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +36 40 – 2a = 10a + 2a +2a 40 = 12a 2-4 Example 2Simplifying Each Side Before Solving Equations

36. Solve. + 1 3 = b – 1 4 = b Can also clear Fractions 2-4 Solve 3x + 15 – 9 = 2(x + 2). 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 x + 6 = 4 – 6 – 6 x = –2

37. Math Joke Q: Why did the variable add its opposite? A: To get to the other side

38. An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. A contradiction is an equation that is not true for any value of the variable. It has no solutions. 2-4Day 2

39. WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2 + 1 = 2 + 1 3 = 3 ALGEBRA 2 + x = 2 + x –x –x 2 = 2 Identities and Contradictions 2-4

40. Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. WORDS x = x + 3 –x 0 = 3  1 = 1 + 2 1 = 3  ALGEBRA NUMBERS Identities and Contradictions 2-4

41. Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x 11 = 11 + 5x + 5x 10 – 5x + 1 = 7x + 11 – 12x All Real or Infinitely Many Solutions 2-4Example 3Infinitely Many Solutions or No Solutions Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –3 = –4 –13x 12x – 3 + x = 5x – 4 + 8x No solution

42. Solve 4y + 7 – y = 10 + 3y. 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 7 = 10 –3y –3y 4y + 7 – y = 10 + 3y  Solve 2c + 7 + c = –14 + 3c + 21. 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 7 = 7 –3c –3c 2c + 7 + c = –14 + 3c + 21 All Real or Infinitely Many Solutions No solution 2-4

43. Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be? PersonBulbs Jon60 bulbs plus 44 bulbs per hour Sara96 bulbs plus 32 bulbs per hour 2-4Example 4Application

44. Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour When is ? 60 + 44b = 96 + 32b – 32b 60 + 12b = 96 –60 – 60 12b = 36 b = 3 hours, 192 Bulbs 2-4

45. Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Let g represent Greg's age, and write expressions for his age. four times Greg's age decreased by 3 is equal to three times Greg's age increased by 7. 4g – 3 = 3g + 7 g – 3 = 7 –3g + 3 + 3 g = 10 Greg is 10 years old. 2-4

46. HW pg. 103 2.4 – Day 1: 15-27 (Odd), 37-43 (Odd), 72, 82, 84 – Day 2: 10-13, 28-36, 14, 52, – Ch: 53, 54, 55, 56

47. 2.5 - Solving for a Variable Algebra I

48. A formula is an equation that states a rule for a relationship among quantities. In the formula d = rt, d is isolated. You can "rearrange" a formula to isolate any variable by using inverse operations. This is called solving for a variable. 2-5Algebra 1 (Bell work)Just Read Solving for a Variable Step 1 Locate the variable you are asked to solve for in the equation. Step 2 Identify the operations on this variable and the order in which they are applied. Step 3 Use inverse operations to undo operations and isolate the variable.

49. The formula C =  d gives the circumference of a circle C in terms of diameter d. The circumference of a bowl is 18 inches. What is the bowl's diameter? Leave the symbol  in your answer. The bowl's diameter is inches. 2-5Example 1Application

50. Solve the formula d = rt for t Find the time in hours that it would take Mr. Mayer to travel 26.2 miles if his average speed was 18 miles per hour. d = rt Mr. Mayer’s time was about 1.46 hours. 2-5

51. The formula for a person’s typing speed is, where s is speed in words per minute, w is number of words typed, e is number of errors, and m is number of minutes typing. Solve for e. 2-5Example 2Solving Formulas for a Variable

52. The formula for the area of a triangle is A = bh, where b is the length of the base, and is the height. Solve for h. A = bh 2-5

53. Math Joke Q: If you give 15 cents to one friend and 10 cents to another friend what time is it? A: A quarter to two 2-5

54. The formula for an object’s final velocity is f = i – gt, where i is the object’s initial velocity, g is acceleration due to gravity, and t is time. Solve for i. f = i – gt + gt +gt f + gt = i 2-5Example 3Solving Literal Equations for a Variable

55. A. Solve x + y = 15 for x. x + y = 15 –y x = –y + 15 B. Solve pq = x for q. pq = x 2-5

56. Solve 5 – b = 2t for t. 5 – b = 2t 2-5 Solve for V VD = m

57. HW pg. 109 2.5 – 2-13, 14-24 (Even), 31, 32, 34 – Ch: 34, 45

58. 2.6 -Solving Absolute Value Equations Algebra I

59. Solve X = 12 Solve 3 X + 7 = 24 2.6

60. Solve -8 = x + 2 - 8 Solve 3 + x + 4 = 0 2.6

61. Math Joke Question: How is the equation x = 8 similar to a chemist's laboratory? Answer: Both have multiple solutions 2.6

62. Solve - 6 + x - 4 = - 6 Solve X - 3 = 4 2.6

63. A support beam for a building must be 3.5 meters long. It is acceptable for the beam to differ from the ideal length by 3 millimeters. Write and solve an absolute-value equation to find the minimum and maximum acceptable lengths for the beam. X – 3.5 =.003 2.6

64. Sydney Harbor Bridge is 134 meters tall. The height of the bridge can rise or fall 180 millimeters because of changes in temperature. Write and Solve an absolute value- equation to find the minimum and maximum heights of the bridge X - 134 = 0.18 2.6

65. HW pg. 115 2.6 – 1-13, 14-28 (Even), 29, 32, 33- 37 (Odd), 44, 47, – B: 43 (Show all work)

66. Algebra I (Bell work) 1.Do Problems # 12, 13, 18, 44 (2.6) 2.If you finish early begin reading 2.7 3.Define: rate, unit rate, dimensional analysis, factor, conversion

67. 2.7 – Rates, Ratios, and Proportions Algebra I

68. 2.7

72. Math Joke Q: What did the circle see when sailing on the ocean? A: Pi - rates 2.7

77. HW pg. 123 2.7- – 1-19, 21-23, 38, 44-47, 70-72 – B: 58

78. Algebra I (Bell work) 1.Do Problems 2, 4, 21, 44 (2.7) (Skip) 2.If you finish early begin reading 2.9 3.Summarize the know it note pg. 133

79. A percent is a ratio that compares a number to 100. For example, To find the fraction equivalent of a percent write the percent as a ratio with a denominator of 100. Then simplify. To find the decimal equivalent of a percent, divide by 100. Just Read 2.9

80. 2.9- Percents Algebra I

81. Some Common Equivalents Percent Fraction Decimal 100% 40% 25% 20%50% 60% 75% 80% 10% 120% = = 1.2 0.5%= = 0.005 Move Decimal 2 places to the left from back 2.9 Just Read

82. Find 30% of 80. Method 1 Use a proportion. 100x = 2400 x = 24 30% of 80 is 24. Find 120% of 15. Method 2 Use an equation. x = 120% of 15 x = 1.20(15) x = 18 120% of 15 is 18. 2.9

83. Find 20% of 60. Method 1 Use a proportion. 100x = 1200 x = 12 20% of 60 is 12. Find 210% of 8. Method 2 Use an equation. x = 210% of 8 x = 2.10(8) x = 17 210% of 8 is 16.8. 2.9

84. What percent of 45 is 35? Round your answer to the nearest tenth. Method 1 Use a proportion. 45x = 3500 x ≈ 77.8 35 is about 77.8% of 45. 2.9

85. 230 is what percent of 200? Method 2 Use a equation. 230 = x 200 230 = 200x 1.15 = x 230 is 115% of 200. 115% = x 2.9

86. What percent of 35 is 7?. Method 1 Use a proportion. 35x = 700 x = 20 7 is 20% of 35. If Time 2.9

87. 27 is what percent of 9? Method 2 Use an equation. 27 = x 9 27 = 9x 3 = x 27 is 300% of 9. 300% = x If Time 2.9

88. 38% of what number is 85? Round your answer to the nearest tenth. Method 1 Use a proportion. 38x = 8500 x = 223.7 38% of about 223.7 is 85. 2.9

89. Math Joke Son: Dad, what does it mean when someone tells me to give 110% Dad: It means they didn’t take Algebra I

90. 20 is 0.4% of what number? Method 2 Use an equation. 20 = 0.4% of x 20 = 0.004 x 5000 = x 20 is 0.4% of 5000. 2.9

91. The serving size of a popular orange drink is 12 oz. The drink is advertised as containing 5% orange juice. How many ounces of orange juice are in one serving size? 100x = 60 x = 0.6 A 12 oz orange drink contains 0.6 oz of orange juice. 2.9

92. HW pg. 136 2.9- – 1-14, 15-25 (Odd), 28-32, 54, 60, – B: 48, 49, 59

93. Algebra I (Bell work) 1.Do Problems # 4, 6, 14 (2.9) 2.If you finish early begin reading 2.10 3.Define: Interest, Principle, Tip, Sales Tax

94. 2.10 Applications of Percents Algebra I

95. Mr. Mayer (Better looking version of Brad Pitt) earns a base salary of $26,000 plus a sales commission of 5%. His total sales for one year were $300,000. Find his total pay for the year. Write the formula for total pay. total pay = base salary + commission Write the formula for commission. Substitute values given in the problem. = 26,000 + 5% of 300,000 = base + % of total sales = 26,000 + (0.05)(300,000) Multiply.= 26,000 + 15,000 Add.= 41,000 Mr. Mayer’s total pay was $41,000. 2.10

96. Write the formula for total pay. total pay = base salary + commission = 350 + 12% of 940 = base + % of total sales = 350 + (0.12)(940) = 350 + 112.80 Add.= 462.80 Dalton Martian total pay was $462.80. Dalton Martian earns $350 per week plus 12% commission on sales. Find his total pay for a week in which his sales were $940. 2.10

97. Interest is the amount of money charged for borrowing money, or the amount of money earned when saving or investing money. Principal is the amount borrowed or invested. Simple interest is interest paid only on the principal. Simple Interest Paid Annually Time in years Interest rate per year as a decimal Principal Simple interest 2.10

98. Math Joke Banker: Do you have an interest on taking out a loan? Customer: If there's’ no interest, I’m interested. 2.10

99. Find the simple interest paid for 3 years on a $2500 loan at 11.5% per year. I = Prt I = (2500)(0.115)(3) I = 862.50 The amount of interest is $862.50. 2.10

100. After 6 months, the simple interest earned on an investment of $5000 was $45. Find the interest rate. I = Prt 45 = 2500r 0.018 = r The interest rate is 1.8%. 2.10

101. The simple interest paid on a loan after 6 months was $306. The annual interest rate was 8%. Find the principal. I = Prt The remaining principal is $7650. 306 = (P)(0.08) 306 =.04P 7650 = P 2.10

102. HW pg.141 2.10- – 2-6, 9-13, 21, 22, 39-45 – B: 26

103. Algebra I (Bell work) 1.Do Problems # 6, 10, 22 (2.10) 2.If you finish early begin reading 2.11 3. Copy the know it note pg. 144, Define: discount/markup

104. 2.11- Percent Increase and Decrease Algebra I

105. A percent change is an increase or decrease given as a percent of the original amount. Percent increase describes an amount that has grown and percent decrease describes an amount that has be reduced. 2.11

106. Find each percent change. Tell whether it is a percent increase or decrease. From 8 to 10 = 0.25 = 25% 8 to 10 is an increase, so a change from 8 to 10 is a 25% increase. 2.11

107. From 75 to 30 = 0.6 = 60% 75 to 30 is a decrease, so a change from 75 to 30 is a 60% decrease. Find the percent change. Tell whether it is a percent increase or decrease. 2.11

108. A. Find the result when 12 is increased by 50%. 0.50(12) = 6 12 + 6 =18 12 increased by 50% is 18. B. Find the result when 55 is decreased by 60%. 0.60(55) = 33 55 – 33 = 22 55 decreased by 60% is 22. 2.11

109. A. Find the result when 72 is increased by 25%. 0.25(72) = 18 72 + 18 =90 72 increased by 25% is 90. B. Find the result when 10 is decreased by 40%. 0.40(10) = 4 10 – 4 = 6 10 decreased by 40% is 6. If Time 2.11

110. Math Joke Q: Why did the shopper think the store was selling everything wholesale? A:Beause the store had two “half-of” sales 2.11

111. Common application of percent change are discounts and markups. A discount is an amount by which an original price is reduced. discount = % of original price final price = original price – discount A markup is an amount by which a wholesale price is increased. final price = wholesale cost markup + wholesale cost = % of 2.11

112. The entrance fee at an amusement park is $35. People over the age of 65 receive a 20% discount. What is the amount of the discount? How much do people over 65 pay? Method 1 A discount is a percent decrease. So find $35 decreased by 20%. 0.20(35) = 7 35 – 7 = 28 Method 2 Subtract the percent discount from 100%. 100% – 20% = 80% 0.80(35) = 28 35 – 28 = 7 By either method, the discount is $7. People over the age of 65 pay $28.00. 2.11

113. A student paid $31.20 for art supplies that normally cost $52.00. Find the percent discount. $52.00 – $31.20 = $20.80 20.80 = x(52.00) 0.40 = x 40% = x The discount is 40% 2.11

114. A video game has a 70% markup. The wholesale cost is $9. What is the selling price? Method 1 A markup is a percent increase. So find $9 increased by 70%. 0.70(9) = 6.30 9 + 6.30 = 15.30 The amount of the markup is $6.30. The selling price is $15.30. 2.11

115. The wholesale cost of a DVD is $7. The markup is 85%. What is the amount of the markup? What is the selling price? Method 1 A markup is a percent increase. So find $7 increased by 85%. 0.85(7) = 5.95 7 + 5.95 = 12.95 12.95  7 = 5.95 By either method, the amount of the markup is $5.95. The selling price is $12.95. Method 2 Add percent markup to 100% 100% + 85% = 185% 1.85(7) = 12.95 2.11

116. What is the percent markup on a car selling for $21,850 that had a wholesale cost of $9500? 21,850 – 9,500 = 12,350 12,350 = x(9,500) 1.30 = x 130% = x The markup was 130 percent. 2.11

117. HW pg. 147 2-11 – 2-15, 32-36 (Even), 49, 50, – B: 51 (Show All Work) – Extra Credit Pg. 152 (Due Friday October 11 th )