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3. Lesson 3-1 Writing Equations
Lesson 3-2 Solving Equations by Using Addition and Subtraction
Lesson 3-3 Solving Equations by Using Multiplication and Division
Lesson 3-4 Solving Multi-Step Equations
Lesson 3-5 Solving Equations with the Variable on Each Side
Lesson 3-6 Ratios and Proportions
Lesson 3-7 Percent of Change
Lesson 3-8 Solving Equations and Formulas
Lesson 3-9 Weighted Averages
Contents

4. Example 1 Translate Sentences into Equations
Example 2 Use the Four-Step Plan
Example 3 Write a Formula
Example 4 Translate Equations into Sentences
Example 5 Write a Problem
Lesson 1 Contents

5. b divided by three is equal to six less than c.
Translate this sentence into an equation.
A number b divided by three is equal to six less than c.
b divided by three is equal to six less than c.
Answer: The equation is .
Example 1-1a

6. 15 z 6 y 2 11 Translate this sentence into an equation.
Fifteen more than z times six is y times two minus eleven.
Fifteen more than z times six is y times two minus eleven. 15 z 6 y 2 11
Answer: The equation is .
Example 1-1b

7. Translate each sentence into an equation.
a. A number c multiplied by six is equal to two more than d.
Answer: The equation is .
b. Three less than a number a divided by four is seven more than 3 times b.
Answer: The equation is .
Example 1-1c

8. Jellybeans A popular jellybean manufacturer produces 1,250,000 jellybeans per hour. How many hours does it take them to produce 10,000,000 jellybeans?
Explore You know that 1,250,000 jellybeans are produced each hour. You want to know how many hours it will take to produce 10,000,000 jellybeans.
Example 1-2a

9. Plan. Write an equation to represent the situation
Plan Write an equation to represent the situation. Let h represent the number of hours needed to produce the jellybeans.
1,250, times hours equals 10,000,000.
1,2500,000 h
10,000,000
Solve
Find h mentally by asking, “What number times 125 equals 1000?”
h = 8
Answer: It will take 8 hours to produce 10,000,000 jellybeans.
Example 1-2b

10. Examine If 1,250,000 jellybeans are produced in one hour, then 1,250,000 x 8 or 10,000,000 jellybeans are produced in 8 hours. The answer makes sense.
Example 1-2c

11. A person at the KeyTronic World Invitational
Type-Off typed 148 words per minute. How
many minutes would it take to type 3552
words?
Answer: It would take 24 minutes.
Example 1-2d

12. P 4s Translate the sentence into a formula.
The perimeter of a square equals four times the length of the side.
Words Perimeter equals four times the length of the side.
Variables Let P = perimeter and s = length of a side.
Perimeter equals four times the length of a side. P 4s
Answer: The formula is .
Example 1-3a

13. Translate the sentence into a formula.
The area of a circle equals the product of  and the square of the radius r.
Answer: The formula is .
Example 1-3b

14. 12 2x 5 Translate this equation into a verbal sentence.
Twelve minus two times x equals negative five.
Answer: Twelve minus two times x equals negative five.
Example 1-4a

15. a squared plus three times b equals c divided by six.
Translate this equation into a verbal sentence.
a b
a squared plus three times b equals c divided by six.
Answer: a squared plus three times b equals c divided by six.
Example 1-4b

16. Translate each equation into a verbal sentence.1.
Answer: Twelve divided by b minus four equals negative one. 2.
Answer: Five times a equals b squared plus one.
Example 1-4c

17. f = cost of fries f + 1.50 = cost of a burger 4( f + 1.50) – f = 8.25
Write a problem based on the given information.
f = cost of fries
f = cost of a burger
4( f ) – f = 8.25
Answer: The cost of a burger is $1.50 more than the cost of fries. Four times the cost of a burger minus the cost of fries equals $8.25. How much do fries cost?
Example 1-5a

18. h = Tiana’s height in inches
Write a problem based on the given information.
h = Tiana’s height in inches
h – 3 = Consuelo’s height in inches
3h(h – 3) = 8262
Answer: Consuelo is 3 inches shorter than Tiana. The product of Consuelo’s height and three times Tiana’s is How tall is Tiana?
Example 1-5b

19. End of Lesson 1

20. Example 1 Solve by Adding a Positive Number
Example 2 Solve by Adding a Negative Number
Example 3 Solve by Subtracting
Example 4 Solve by Adding or Subtracting
Example 5 Write and Solve an Equation
Example 6 Write an Equation to Solve a Problem
Lesson 2 Contents

21. Solve . Then check your solution.
Original equation
Add 12 to each side.
Answer:
and
To check that –15 is the solution, substitute –15 for h in the original equation.
Example 2-1a

22. Solve . Then check your solution.
Answer: 40
Check:
Example 2-1b

23. Solve . Then check your solution.
Original equation
Add –63 to each side.
Answer:
and
To check that 29 is the solution, substitute 29 for k in the original equation.
Example 2-2a

24. Solve . Then check your solution.
Answer: –61
Check:
Example 2-2b

25. Solve . Then check your solution.
Original equation
Subtract 102 from each side.
Answer:
and
To check that –66 is the solution, substitute –66 for c in the original equation.
Example 2-3a

26. Solve . Then check your solution.
Answer: –171
Check:
Example 2-3b

27. Method 1 Use the Subtraction Property of Equality.
Solve in two ways.
Method 1 Use the Subtraction Property of Equality.
Original equation
Subtract from each side.
and
Answer: or
Example 2-4a

28. Method 2 Use the Addition Property of Equality.
Original equation
Add to each side.
and
Answer: or
Example 2-4b

29. Solve .
Answer:
Example 2-4c

30. Write an equation for the problem
Write an equation for the problem. Then solve the equation and check your solution.
Fourteen more than a number is equal to twenty-seven. Find this number.
Fourteen more than a number is equal to twenty-seven. 14 n 27
Original equation
Subtract 14 from each side.
Example 2-5a

31. and
Answer:
To check that 13 is the solution, substitute 13 for n in the original equation. The solution is 13.
Example 2-5b

32. Twelve less than a number is equal to negative twenty-five
Twelve less than a number is equal to negative twenty-five. Find the number.
Answer: –13
Example 2-5c

33. History The Washington Monument in Washington, D. C
History The Washington Monument in Washington, D.C., was built in two phases. From 1848–1854, the monument was built to a height of 152 feet. From 1854 until 1878, no work was done. Then from 1878 to 1888, the additional construction resulted in its final height of 555 feet. How much of the monument was added during the second construction phase?
Example 2-6a

34. Words The first height plus the additional height equals 555 feet.
Variables Let a = the additional height.
The first height plus the additional height equals
152 a
555
Original equation
Subtract 152 from each side.
and
Answer: There were 403 feet added to the Washington Monument from 1878 to 1888.
Example 2-6b

35. The Sears Tower was built in 1974
The Sears Tower was built in The height to the Sky Deck is 1353 feet. The actual recorded height is 1450 feet. In 1982, they added twin antenna towers, which does not count for the record, for a total structure height of 1707 feet. How tall are the twin antenna towers?
Answer: 257 feet
Example 2-6c

36. End of Lesson 2

37. Example 1 Solve Using Multiplication by a Positive Number
Example 2 Solve Using Multiplication by a Fraction
Example 3 Solve Using Multiplication by a Negative Number
Example 4 Write and Solve an Equation Using Multiplication
Example 5 Solve Using Division by a Positive Number
Example 6 Solve Using Division by a Negative Number
Example 7 Write and Solve an Equation Using Division
Lesson 3 Contents

38. Solve . Then check your solution.
Original equation
Multiply each side by 12.
Answer:
and
To check that 9 is the solution, substitute 9 for s in the original equation.
Example 3-1a

39. Solve .
Answer: 12
Example 3-1b

40. Solve . Then check your solution.
Original equation
Rewrite each mixed number as in improper fraction.
Multiply each side by , the reciprocal of
Example 3-2a

41. Answer: or Check this result.
To check that is the solution, substitute for k in the original equation.
Example 3-2b

42. Solve .
Answer:
Example 3-2c

43. Multiply each side by , the reciprocal of –15.
Solve .
Original equation
Multiply each side by , the reciprocal of –15.
Answer:
Check this result.
To check that 5 is the solution, substitute 5 for b in the original equation.
Example 3-3a

44. Solve .
Answer:
Example 3-3b

45. Words Three times the weight on Mars equals the weight on Earth.
Space Travel Using information from Example 4 in the Student Edition, what would be the weight of Neil Armstrong’s suit and life-support backpack on Mars if three times the Mars weight equals the Earth weight?
Words Three times the weight on Mars equals the weight on Earth.
Variables Let w = the weight on Mars.
Example 3-4a

46. 3 w 198 Original equation Multiply each side by . and
Three times the weight on Mars equals the weight on Earth. 3 w
198
Original equation
Multiply each side by .
and
Answer: The weight of Neil Armstrong’s suit and life-support backpacks on Mars would be 66 pounds.
Example 3-4b

47. Answer: He would weigh 72 pounds on Mars.
Refer to the information about Neil Armstrong in Example 4. If Neil Armstrong weighed 216 pounds on Earth, how much would he weigh on Mars?
Answer: He would weigh 72 pounds on Mars.
Example 3-4c

48. Solve . Then check your solution.
Original equation
Divide each side by 11.
Answer:
and
To check, substitute 13 for w.
Example 3-5a

49. Solve . Then check your solution.
Answer: 17
Check:
Example 3-5b

50. Solve . Original equation Divide each side by –8. and Answer:
Example 3-6a

51. Solve .
Answer: –23
Example 3-6b

52. Write an equation for the problem below. Then solve the equation.
Negative fourteen times a number equals 224.
Negative fourteen times a number equals n
–14
224
Original equation
Divide each side by –14.
Answer:
Check this result.
Example 3-7a

53. Negative thirty-four times a number equals 578. Find the number.
Answer: –17
Example 3-7b

54. End of Lesson 3

55. Example 1 Work Backward to Solve a Problem
Example 2 Solve Using Addition and Division
Example 3 Solve Using Subtraction and Multiplication
Example 4 Solve Using Multiplication and Addition
Example 5 Write and Solve a Multi-Step Equation
Example 6 Solve a Consecutive Integer Problem
Lesson 4 Contents

56. Danny took some rope with him on his camping trip
Danny took some rope with him on his camping trip. He used 32 feet of rope to tie his canoe to a log on the shore. The next night, he used half of the remaining rope to secure his tent during a thunderstorm. On the last day, he used 7 feet as a fish stringer to keep the fish that he caught. After the camping trip, he had 9 feet left. How much rope did he have at the beginning of the camping trip?
Example 4-1a

57. Start at the end of the problem and undo each step.
Statement Undo the Statement
He had 9 feet left. 9
He used 7 feet as a fish stringer.
9 + 7 = 16
He used half of the remaining rope to secure his tent.
16  2 = 32
He used 32 feet to tie his canoe.
= 64
Answer: He had 64 feet of rope. Check the answer in the context of the problem.
Example 4-1b

58. Olivia went to the mall to spend some of her monthly allowance
Olivia went to the mall to spend some of her monthly allowance. She put $10 away so it could be deposited in the savings account at a later date. The first thing she bought was a CD for $ The next stop was to buy hand lotion and a candle, which set her back $9.59. For lunch, she spent half of the remaining cash. She went to the arcade room and spent $5.00 and took home $1.21. How much was Olivia’s monthly allowance?
Answer: $48.00
Example 4-1c

59. Solve . Then check your solution.
Original equation
Add 13 to each side.
Simplify.
Divide each side by 5.
Answer:
Simplify.
To check, substitute 10 for q in the original equation.
Example 4-2a

60. Solve .
Answer: 14
Example 4-2b

61. Solve . Then check your solution.
Original equation
Subtract 9 from each side.
Simplify.
Multiply each side by 12.
Answer: s = –240
Simplify.
Example 4-3a

62. To check, substitute –240 for s in the original equation.
Example 4-3b

63. Solve .
Answer: 363
Example 4-3c

64. Solve . Original equation Multiply each side by –3. Simplify.
Add 8 to each side.
Answer: r = 14
Simplify.
Example 4-4a

65. Solve .
Answer: 21
Example 4-4b

66. Write an equation for the problem below. Then solve the equation.
Eight more than five times a number is negative 62.
Eight more than five times a number is negative 62. n 8 62 5
Original equation
Subtract 8 from each side.
Simplify.
Example 4-5a

67. Multiply each side by .
Simplify.
Answer: n = –14
Example 4-5b

68. Three-fourths of seven subtracted from a number is negative fifteen
Three-fourths of seven subtracted from a number is negative fifteen. What is the number?
Answer: –13
Example 4-5c

69. Number Theory Write an equation for the problem below
Number Theory Write an equation for the problem below. Then solve the equation and answer the problem.
Find three consecutive odd integers whose sum is 57.
Let n = the least odd integer.
Let n + 2 = the next greater odd integer.
Let n + 4 = the greatest of the three odd integers.
The sum of three consecutive odd integers is = 57
Example 4-6a

70. Subtract 6 from each side.
Original equation
Simplify.
Subtract 6 from each side.
Simplify.
Divide each side by 3.
Simplify.
or 19
Answer: The consecutive odd integers are 17, 19, and 21.
or 21
Example 4-6b

71. Find three consecutive even integers whose sum is 84.
Answer: 26, 28, 30
Example 4-6c

72. End of Lesson 4

73. Example 1 Solve an Equation with Variables on Each Side
Example 2 Solve an Equation with Grouping Symbols
Example 3 No Solutions
Example 4 An Identity
Example 5 Use Substitution to Solve an Equation
Lesson 5 Contents

74. Subtract 7s from each side.
Solve .
Original equation
Subtract 7s from each side.
Simplify.
Subtract 8 from each side.
Simplify.
Divide each side by –2.
Answer: s = 5
Simplify.
Example 5-1a

75. To check your answer, substitute 5 for s in the original equation.
Example 5-1b

76. Solve .
Answer:
Example 5-1c

77. Distributive Property
Solve . Then check your solution.
Original equation
Distributive Property
Subtract 12q from each side.
Simplify.
Subtract 6 from each side.
Example 5-2a

78. To check, substitute 6 for q in the original equation.
Simplify.
Divide each side by –8.
Answer: q = 6
Simplify.
To check, substitute 6 for q in the original equation.
Example 5-2b

79. Solve .
Answer: 36
Example 5-2c

80. Distributive Property
Solve .
Original equation
Distributive Property
Subtract 40c from each side.
This statement is false.
Answer: There must be at least one c to represent the variable. This equation has no solution.
Example 5-3a

81. Answer: This equation has no solution.
Solve .
Answer: This equation has no solution.
Example 5-3b

82. Solve . Then check your solution.
Original equation
Distributive Property
Answer: Since the expression on each side of the equation is the same, this equation is an identity. The statement is true for all values of t.
Example 5-4a

83. Answer: is true for all values of c.
Solve .
Answer: is true for all values of c.
Example 5-4b

84. Multiple-Choice Test Item Solve . A 13 B –13 C 26 D –26
Read the Test Item You are asked to solve an equation.
Solve the Test Item You can solve the equation or substitute each value into the equation and see if it makes the equation true. We will solve by substitution.
Example 5-5a

85. A: Substitute 13 for b.
Example 5-5b

86. B: Substitute –13 for b.
Example 5-5c

87. C: Substitute 26 for b.
Example 5-5d

88. Answer: Since the value –26 makes the statement true, the answer is D.
D: Substitute –26 for b.
Answer: Since the value –26 makes the statement true, the answer is D.
Example 5-5e

89. Multiple-Choice Test Item Solve . A 32 B –32 C 26 D –26
Answer: C
Example 5-5f

90. End of Lesson 5

91. Example 1 Determine Whether Ratios Form a Proportion
Example 2 Use Cross Products
Example 3 Solve a Proportion
Example 4 Use Rates
Example 5 Use a Scale Drawing
Lesson 6 Contents

92. Determine whether the ratios and form a proportion.
Answer: The ratios are equal. Therefore, they form a proportion.
Example 6-1a

93. Do the ratios and form a proportion?
Answer: The ratios are not equal. Therefore, they do not form a proportion.
Example 6-1b

94. Find the cross products.
Use cross products to determine whether the pair of ratios below forms a proportion.
Write the equation.
Find the cross products.
Simplify.
Answer: The cross products are not equal, so . The ratios do not form a proportion.
Example 6-2a

95. Find the cross products.
Use cross products to determine whether the pair of ratios below forms a proportion.
Write the equation.
Find the cross products.
Simplify.
Answer: The cross products are equal, so . Since the ratios are equal, they form a proportion.
Example 6-2b

96. Use cross products to determine whether the pair of ratios below forms a proportion.
Answer: The cross products are equal. Therefore, the ratios do form a proportion.
Example 6-2c

97. Use cross products to determine whether the pair of ratios below forms a proportion.
Answer: The cross products are not equal. Therefore, the ratios do not form a proportion.
Example 6-2d

98. Find the cross products.
Solve the proportion .
Original equation
Find the cross products.
Simplify.
Divide each side by 8.
Answer:
Simplify.
Example 6-3a

99. Solve the proportion .
Answer: 6.3
Example 6-3b

100. Plan Write a proportion for the problem.
Bicycling The gear on a bicycle is 8:5. This means that for every eight turns of the pedals, the wheel turns five times. Suppose the bicycle wheel turns about times during a trip. How many times would you have to turn the pedals during the trip?
Explore Let p represent the number of times needed to crank the pedals.
Plan Write a proportion for the problem.
turns of the pedals
wheel turns
Example 6-4a

101. Find the cross products.
Solve
Original proportion
Find the cross products.
Simplify.
Divide each side by 5.
Answer: 3896 = p
Simplify.
Example 6-4b

102. Examine If it takes 8 turns of the pedal to make the wheel turn 5 times, then it would take 1.6 turns of the pedal to make the wheel turn 1 time. So, if the wheel turns 2435 times, then there are 2435  1.6 or 3896 turns of the pedal. The answer is correct.
Example 6-4c

103. Answer: About 60,480 cels were drawn to produce Snow White.
Before 1980, Disney created animated movies using cels. These hand drawn cels (pictures) of the characters and scenery represented the action taking place, one step at a time. For the movie Snow White, it took 24 cels per second to have the characters move smoothly. The movie is around 42 minutes long. About how many cels were drawn to produce Snow White?
Answer: About 60,480 cels were drawn to produce Snow White.
Example 6-4d

104. Explore Let d represent the actual distance.
Maps In a road atlas, the scale for the map of Connecticut is 5 inches = 41 miles. The scale for the map of Texas is 5 inches = 144 miles. What are the distances in miles represented by 2.5 inches on each map?
Explore Let d represent the actual distance.
Plan Write a proportion for the problem.
scale
actual
Connecticut:
Example 6-5a

105. Texas:
scale
actual
Example 6-5b

106. Find the cross products.
Solve Connecticut:
Find the cross products.
Simplify.
Divide each side by 5.
Simplify.
or 20.5
Example 6-5c

107. Find the cross products.
Solve Texas:
Find the cross products.
Simplify.
Divide each side by 5.
Simplify.
or 72
Example 6-5d

108. Answer:. The actual distance in Connecticut represented by 2
Answer: The actual distance in Connecticut represented by 2.5 inches is 20.5 miles. The actual distance in Texas represented by 2.5 inches is 72 miles.
Examine: 2.5 inches is of 5 inches. So 2.5 inches represents (41) or 20.5 miles in Connecticut and (144) or 72 miles in Texas. The answer is correct.
Example 6-5f

109. The scale on a map of the United States is. inches = 750 miles
The scale on a map of the United States is inches = 750 miles. The distance, on the map, between Los Angeles and Washington, D.C., is about inches. What is the distance in miles between the two locations?
Answer: The distance in miles between Los Angeles and Washington, D.C., is about 2,114 miles.
Example 6-5g

110. End of Lesson 6

111. Example 1 Find Percent of Change Example 2 Find the Missing Value
Example 3 Find Amount After Sales Tax
Example 4 Find Amount After Discount
Lesson 7 Contents

112. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
original: 32 new: 40
Find the amount of change. Since the new amount is greater than the original, the percent of change is a percent of increase.
40 – 32 = 8
Example 7-1a

113. Find the percent using the original number, 32, as the base.
change
original amount
percent change
100 percent
Find the cross products.
Simplify.
Divide each side by 32.
Simplify.
Answer: The percent of increase is 25%.
Example 7-1b

114. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
original: 20 new: 4
Find the amount of change. Since the new amount is less than the original, the percent of change is a percent of decrease.
20 – 4 = 16
Example 7-1c

115. Find the percent using the original number, 20, as the base.
change
original amount
percent change
100 percent
Find the cross products.
Simplify.
Divide each side by 20.
Simplify.
Answer: The percent of decrease is 80%.
Example 7-1d

116. Answer: The percent of change is a decrease of 28%.
a. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
original: 20 new: 18
Answer: The percent of change is a decrease of 28%.
Example 7-1e

117. Answer: The percent of change is an increase of 300%.
b. State whether the percent of change is a percent of increase or a percent of decrease. Then find the percent of change.
original: 12 new: 48
Answer: The percent of change is an increase of 300%.
Example 7-1f

118. Sales The price a used-book store pays to buy a book is $5
Sales The price a used-book store pays to buy a book is $5. The store sells the book for 28% above the price that it pays for the book. What is the selling price of the $5 book?
Let s = the selling price of the book. Since 28% is the percent of increase, the amount the used-book store pays to buy a book is less than the selling price. Therefore, s – 5 represents the amount of change.
Example 7-2a

119. Find the cross products.
change
book store cost
percent change
100 percent
Find the cross products.
Distributive Property
Add 500 to each side.
Simplify.
Divide each side by 100.
Simplify.
Answer: The selling price of the $5 book is $6.40.
Example 7-2b

120. Answer: The price of jeans at the second store is $31.72.
At one store the price of a pair of jeans is $ At another store the same pair of jeans has a price that is 22% higher. What is the price of jeans at the second store?
Answer: The price of jeans at the second store is $31.72.
Example 7-2c

121. The tax is 5% of the price of the meal.
Sales Tax A meal for two at a restaurant costs $ If the sales tax is 5%, what is the total price of the meal?
The tax is 5% of the price of the meal.
Use a calculator.
Round $ to $1.64. Add this amount to the original price.
Answer: The total price of the meal is $34.49.
Example 7-3a

122. Answer: The total price of the CD player is $74.71.
A portable CD player costs $ If the sales tax is 6.75%, what is the total price of the CD player?
Answer: The total price of the CD player is $74.71.
Example 7-3b

123. The discount is 20% of the original price.
Discount A dog toy is on sale for 20% off the original price. If the original price of the toy is $3.80, what is the discounted price?
The discount is 20% of the original price.
Subtract $0.76 from the original price.
Answer: The discounted price of the dog toy is $3.04.
Example 7-4a

124. Answer: The discounted price of the cap player is $16.99.
A baseball cap is on sale for 15% off the original price. If the original price of the cap is $19.99, what is the discounted price?
Answer: The discounted price of the cap player is $16.99.
Example 7-4b

125. End of Lesson 7

126. Example 1 Solve an Equation for a Specific Variable
Example 3 Use a Formula to Solve Problems
Example 4 Use Dimensional Analysis
Lesson 8 Contents

127. Subtract 12c from each side.
Solve for b.
Original equation
Subtract 12c from each side.
Simplify.
Divide each side by 5.
Example 8-1a

128. Answer: The value of b is .or
Simplify.
Answer: The value of b is .
Example 8-1b

129. Answer: The value of y is .
Solve for y.
Answer: The value of y is .
Example 8-1c

130. Use the Distributive Property.
Solve for x.
Original equation
Add xy to each side.
Simplify.
Add 2z to each side.
Simplify.
Use the Distributive Property.
Example 8-2a

131. Answer: The value of x is . Since division by 0 is undefined, .
Divide each side by 7 + y.
Answer: The value of x is . Since division by 0 is
undefined, .
Example 8-2b

132. Answer: The value of a is .
Solve for a.
Answer: The value of a is .
Example 8-2c

133. Fuel Economy A car’s fuel economy E (miles per gallon) is given by the formula , where m is the number of miles driven and g is the number of gallons of fuel used. Solve the formula for m.
Example 8-3a

134. Formula for fuel economy.
Multiply each side by g.
Simplify.
Answer:
Example 8-3b

135. Fuel Economy If Claudia’s car has an average fuel consumption of 30 miles per gallon and she used 9.5 gallons, how far did she drive?
Formula for how many miles driven
E = 30 mpg and g = 9.5 gallons
Multiply.
Answer: She drove 285 miles.
Example 8-3c

136. Fuel Economy A car’s fuel economy E (miles per gallon) is given by the formula , where m is the number of miles driven and g is the number of gallons of fuel used. Solve the formula for g.
Answer:
Example 8-3d

137. Answer: She used around 77.74 gallons.
If Claudia drove 1477 miles and her pickup has an average fuel consumption of 19 miles per gallon, how many gallons of fuel did she use?
Answer: She used around gallons.
Example 8-3e

138. Geometry The formula for the volume of a cylinder is
Geometry The formula for the volume of a cylinder is , where r is the radius of the cylinder and h is the height. Solve the formula for h.
Original formula
Divide each side by .
Answer:
Example 8-4a

139. Geometry What is the height of a cylindrical swimming pool that has a radius of 12 feet and a volume of 1810 cubic feet?
Formula for h
V = 1810 and r = 12
Use a calculator.
Answer: The height of the cylindrical swimming pool is about 4 feet.
Example 8-4b

140. Geometry The formula for the volume of a cylinder is
Geometry The formula for the volume of a cylinder is , where r is the radius of the cylinder and h is the height.
Solve the formula for r.
Answer:
Example 8-4c

141. Answer: The radius is about 10 feet.
What is the radius of a cylindrical swimming pool if the volume is 2010 cubic feet and a height of 6 feet?
Answer: The radius is about 10 feet.
Example 8-4d

142. End of Lesson 8

143. Example 1 Solve a Mixture Problem with Prices
Example 2 Solve a Mixture Problem with Percents
Example 3 Solve for Average Speed
Example 4 Solve a Problem Involving Speeds of Two Vehicles
Lesson 9 Contents

144. Pets Jeri likes to feed her cat gourmet cat food that costs $1
Pets Jeri likes to feed her cat gourmet cat food that costs $1.75 per pound. However, food at that price is too expensive so she combines it with cheaper cat food that costs $0.50 per pound. How many pounds of cheaper food should Jeri buy to go with 5 pounds of gourmet food, if she wants the price to be $1.00 per pound?
Example 9-1a

145. Let w = the number of pounds of cheaper cat food. Make a table.
Units (lb)
Price per Unit
Price
Gourmet cat food
Mixed cat food 5
$1.75
$8.75 w
$0.50
0.5w
5 + w
$1.00
1.00(5 + w)
Example 9-1b

146. Price of gourmet cat food price of cheaper cat food
Write and solve an equation using the information in the table.
Price of gourmet cat food
plus
price of cheaper cat food
equals
price of mixed cat food.
8.75
0.5w
1.00(5 + w)
Original equation
Distributive Property
Subtract 0.5w from each side.
Simplify.
Example 9-1c

147. Subtract 5.0 from each side.
Simplify.
Divide each side by 0.5.
Simplify.
Answer: Jerry should buy 7.5 pounds of cheaper cat food to be mixed with the 4 pounds of gourmet cat food to equal out to $1.00 per pound of cat food.
Example 9-1d

148. Answer: Cheryl should buy 4.6 ounces of beads.
Cheryl has bought 3 ounces of sequins that cost $1.79 an ounce. The seed beads cost $0.99 an ounce. How many ounces of seed beads can she buy if she only wants the beads to be $1.29 an ounce for her craft project?
Answer: Cheryl should buy 4.6 ounces of beads.
Example 9-1e

149. Auto Maintenance To provide protection against freezing, a car’s radiator should contain a solution of 50% antifreeze. Darryl has 2 gallons of a 35% antifreeze solution. How many gallons of 100% antifreeze should Darryl add to his solution to produce a solution of 50% antifreeze?
Example 9-2a

150. Amount of Solution (gallons)
Let g = the number of gallons of 100% antifreeze to be added. Make a table.
35% Solution
100% Solution
50% Solution
Amount of Solution (gallons)
Price 2
0.35(2) g
1.0(g)
2 + g
0.50(2 + g)
Example 9-2b

151. Write and solve an equation using the information in the table.
Amount of antifreeze in 35% solution
plus
amount of antifreeze in 100% solution
equals
amount of antifreeze in 50% solution.
0.35(2)
1.0(g)
0.50(2 + g)
Original equation
Distributive Property
Subtract 0.50g from each side.
Simplify.
Example 9-2c

152. Subtract 0.70 from each side.
Simplify.
Divide each side by 0.50.
Simplify.
Answer: Darryl should add 0.60 gallons of 100% antifreeze to produce a 50% solution.
Example 9-2d

153. Answer: of a pound of 75% peanuts should be used.
A recipe calls for mixed nuts with 50% peanuts pound of 15% peanuts has already been used. How many pounds of 75% peanuts needs to be add to obtain the required 50% mix?
Answer: of a pound of 75% peanuts should be used.
Example 9-2e

154. To find the average speed for each leg of the trip, rewrite .
Air Travel Mirasol took a non-stop flight from Newark to Austin to visit her grandmother. The 1500-mile trip took three hours and 45 minutes. Because of bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip?
To find the average speed for each leg of the trip, rewrite .
Example 9-3a

155. Going
Returning
Example 9-3b

156. Definition of weighted average Round Trip
Simplify.
Answer: The average speed for the round trip was about miles per hour.
Example 9-3c

157. In the morning, when traffic is light, it takes 30 minutes to get to work. The trip is 15 miles through towns. In the afternoon when traffic is a little heavier, it takes 45 minutes. What is the average speed for the round trip?
Answer: The average speed for the round trip was about 23 miles per hour.
Example 9-3d

158. Rescue A railroad switching operator has discovered that two trains are heading toward each other on the same track. Currently, the trains are 53 miles apart. One train is traveling at 75 miles per hour and the other train is traveling at 40 miles per hour. The faster train will require 5 miles to stop safely, and the slower train will require 3 miles to stop safely. About how many minutes does the operator have to warn the train engineers to stop their trains?
Example 9-4a

159. Draw a diagram. 53 miles apart Takes 5 miles to stop
Example 9-4b

160. Let m = the number of minutes that the operator has to warn the train engineers to stop their trains safely. Make a table.
Fast train
Other train r
d = rt t 75 m
75m 40 m
40m
Example 9-4c

161. Distance traveled by fast train distance traveled by other train
Write and solve an equation using the information in the table.
Distance traveled by fast train
plus
distance traveled by other train
equals
45 miles.
75m
40m 45
Original equation
Simplify.
Divide each side by 115.
Example 9-4d

162. Round to the nearest hundredth.
Convert to minutes by multiplying by 60.
Answer: The operator has about 23 minutes to warn the engineers.
Example 9-4e

163. Answer: They will be 7.5 miles apart in about 14 minutes.
Two students left the school on their bicycles at the same time, one heading north and the other heading south. The student heading north travels 15 miles per hour, and the one heading south travels at 17 miles per hour. About how many minutes will they be 7.5 miles apart?
Answer: They will be 7.5 miles apart in about 14 minutes.
Example 9-4f

164. End of Lesson 9

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