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- 1. Proportions Frank Ma © 2011
- 2. Proportions Two related quantities stated side by side is called a ratio.
- 3. Proportions Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4.
- 4. 3 4 Proportions Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 5. 3 4 eggs cups of flour Proportions Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 6. 3 4 eggs cups of flour Proportions This fraction is also the amount of per unit of the given ratio, in this case, ¾ egg / per cup of flour. Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 7. 3 4 eggs cups of flour Proportions This fraction is also the amount of per unit of the given ratio, in this case, ¾ egg / per cup of flour. Two ratios that are equal are said to be in proportion. Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 8. 3 4 eggs cups of flour Proportions This fraction is also the amount of per unit of the given ratio, in this case, ¾ egg / per cup of flour. Two ratios that are equal are said to be in proportion. Thus "3 to 4" is proportion to "6 to 8" since 3 4 = 6 8 Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 9. 3 4 eggs cups of flour Proportions This fraction is also the amount of per unit of the given ratio, in this case, ¾ egg / per cup of flour. Two ratios that are equal are said to be in proportion. Thus "3 to 4" is proportion to "6 to 8" since 3 4 = 6 8 Proportional equations are the simplest type of fractional equations. Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 10. 3 4 eggs cups of flour Proportions This fraction is also the amount of per unit of the given ratio, in this case, ¾ egg / per cup of flour. Two ratios that are equal are said to be in proportion. Thus "3 to 4" is proportion to "6 to 8" since 3 4 = 6 8 Proportional equations are the simplest type of fractional equations. To solve proportional equations, we cross-multiply and change the proportions into regular equations. Two related quantities stated side by side is called a ratio. For example, if a recipe calls for 3 eggs and 4 cups of flour, then the ratio of eggs to flour is 3 to 4. We may write it using fractional notation as:
- 11. A B C D , =If Cross-Multiplication-Rule Proportions
- 12. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions
- 13. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a.
- 14. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply
- 15. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x
- 16. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x5 6
- 17. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x 2 3 (x + 2) (x – 5) =b. 5 6
- 18. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x5 6 2 3 (x + 2) (x – 5) =b. cross multiply
- 19. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x 2 3 (x + 2) (x – 5) =b. cross multiply 2(x – 5) = 3(x + 2) 5 6
- 20. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x 2 3 (x + 2) (x – 5) =b. cross multiply 2(x – 5) = 3(x + 2) 2x – 10 = 3x + 6 5 6
- 21. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x 2 3 (x + 2) (x – 5) =b. cross multiply 2(x – 5) = 3(x + 2) 2x – 10 = 3x + 6 –10 – 6 = 3x – 2x 5 6
- 22. A B C D , =If then AD = BC. Cross-Multiplication-Rule Proportions Example A. Solve for x. 3 x 5 2 =a. cross multiply 6 = 5x = x 2 3 (x + 2) (x – 5) =b. cross multiply 2(x – 5) = 3(x + 2) 2x – 10 = 3x + 6 –10 – 6 = 3x – 2x –16 = x 5 6
- 23. Proportions When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions
- 24. Proportions When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 25. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 26. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 27. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 28. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. number of eggs cups of flour When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 29. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 30. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour x 10 When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 31. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour 3 4 x 10 When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 32. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour 3 4 x 10 = Set them equal, we get x 10 3 4 When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 33. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour 3 4 x 10 = Set them equal, we get x 10 3 4 cross multiply 4x = 30 When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 34. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour 3 4 x 10 = Set them equal, we get x 10 3 4 cross multiply 4x = 30 x = 30 4 When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 35. Proportions Example B. A recipe calls for 3 eggs and 4 cups of flour. How many eggs are needed if we have 10 cups of flour? Let x be the number of eggs needed. Write the two types of measurements, the number of eggs and the number of cups of flour, in a column as shown. Then write down the ratios. number of eggs cups of flour 3 4 x 10 = Set them equal, we get x 10 3 4 cross multiply 4x = 30 x = 30 4 = 7½ We need 7½ eggs. When setting up a proportional equation for a word problem, the two quantities for the same type of measurement must occupy the same position in the fractions i.e. both must be in numerator, or both must be in the denominator.
- 36. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions
- 37. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as
- 38. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches
- 39. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches x 14
- 40. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches 21 4 x 14
- 41. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches 21 4 x 14 Set them equal, we get 21 4 x 14 =
- 42. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches 21 4 x 14 Set them equal, we get 21 4 x 14 = cross multiply 4x = 294
- 43. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches 21 4 x 14 Set them equal, we get 21 4 x 14 = cross multiply 4x = 294 x = 294 4
- 44. Example C. On a map, 4 inches corresponds to 21 miles in real distance. What is the real distance between two points if they are 14 inches apart on the map? Proportions Let x be the number of real distance in miles and write the two types of measurements as miles inches 21 4 x 14 Set them equal, we get 21 4 x 14 = cross multiply 4x = 294 x = 294 4 = 73½ So 4 inches corresponds to 73½ miles.
- 45. Proportions Geometric proportion gives us the sense of “similarity”, that is, objects that are of the same shape but different sizes.
- 46. Proportions Given the following cats, which cat on the right is of the same shape as the one on the left? A. B. Geometric proportion gives us the sense of “similarity”, that is, objects that are of the same shape but different sizes.
- 47. Proportions Given the following cats, which cat on the right is of the same shape as the one on the left? The answer of course is A. A. B. Geometric proportion gives us the sense of “similarity”, that is, objects that are of the same shape but different sizes.
- 48. Proportions Given the following cats, which cat on the right is of the same shape as the one on the left? The answer of course is A. A. B. This sense of similarity is due to the fact that “all corresponding linear measurements are in proportion”. Geometric proportion gives us the sense of “similarity”, that is, objects that are of the same shape but different sizes.
- 49. Proportions For example, given the following similar cats,
- 50. Proportions For example, given the following similar cats, let’s identify some corresponding points as shown.
- 51. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. 6 10 12
- 52. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. 6 10 12 3
- 53. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. Then we may find the distances y and z by proportion. 6 10 12 3 y z
- 54. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. Then we may find the distances y and z by proportion. Since the distances between the tips of the ears is 6 : 3 or 2 : 1 ratio, 6 10 12 3 y z
- 55. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. Then we may find the distances y and z by proportion. Since the distances between the tips of the ears is 6 : 3 or 2 : 1 ratio, we see that y = 6 10 12 3 y z
- 56. Proportions B. For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. Then we may find the distances y and z by proportion. Since the distances between the tips of the ears is 6 : 3 or 2 : 1 ratio, we see that y = 10/5 = 2 and z = 12/2 = 6. 6 10 12 3 y z
- 57. Proportions 6 10 12 3 5 6 For example, given the following similar cats, let’s identify some corresponding points as shown. Suppose the following measurements are known. Then we may find the distances y and z by proportion. Since the distances between the tips of the ears is 6 : 3 or 2 : 1 ratio, we see that y = 10/5 = 2 and z = 12/2 = 6.
- 58. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B.
- 59. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B. Hence if X, Y and Z are the measurements from the original cat as shown, X Y Z
- 60. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B. Hence if X, Y and Z are the measurements from the original cat as shown, and x, y and z are the corresponding measurements in the similar copy, X Y Z x y z
- 61. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B. Hence if X, Y and Z are the measurements from the original cat as shown, and x, y and z are the corresponding measurements in the similar copy, then it must be that X Y Z x y z X x = Y y = Z z
- 62. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B. Hence if X, Y and Z are the measurements from the original cat as shown, and x, y and z are the corresponding measurements in the similar copy, then it must be that X Y Z x y z X x = Y y = Z z or that X Y = x y etc…
- 63. Proportions Definition. Two geometric objects are similar if all corresponding linear measurements are in proportion. B. Hence if X, Y and Z are the measurements from the original cat as shown, and x, y and z are the corresponding measurements in the similar copy, then it must be that X Y Z x y z X x = Y y = Z z or that X Y = x y etc… This mathematical formulation of similarity is the basis for biometric security software such as facial recognition systems.
- 64. Proportions The reason that linear proportionality gives us the sense of similarity is due to visual geometry, that is, the visual images of an object at different distances form a “cone”.
- 65. Proportions The reason that linear proportionality gives us the sense of similarity is due to visual geometry, that is, the visual images of an object at different distances form a “cone”.
- 66. Proportions The reason that linear proportionality gives us the sense of similarity is due to visual geometry, that is, the visual images of an object at different distances form a “cone”. One can easily demonstrate that in such projection, similar images of two different sizes must preserve the ratio of two corresponding distance measurements – as in the above definition of similarity.
- 67. Proportions The reason that linear proportionality gives us the sense of similarity is due to visual geometry, that is, the visual images of an object at different distances form a “cone”. One can easily demonstrate that in such projection, similar images of two different sizes must preserve the ratio of two corresponding distance measurements – as in the above definition of similarity. Similar triangles
- 68. Proportions In the definition of similarity, the measurements in proportion must be linear.
- 69. Proportions In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects.
- 70. Proportions In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, 1 2
- 71. Proportions In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the area–measurements is 1 : (2)2 or 1 : 4 ratio. 1 2 linear ratio 1 : 2 area–ratio 1 : 4 = 22 :
- 72. Proportions 1 2 linear ratio 1 : 2 area–ratio 1 : 4 = 22 : In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the area–measurements is 1 : (2)2 or 1 : 4 ratio. So it would take 4 times as much paint to paint the larger one.
- 73. Proportions 1 2 linear ratio 1 : 2 area–ratio 1 : 4 = 22 linear ratio 1 : r area–ratio 1 : r2 1 r : : In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the area–measurements is 1 : (2)2 or 1 : 4 ratio. So it would take 4 times as much paint to paint the larger one. Given two similar objects with linear ratio 1 : r then their area–ratio is 1 : r2.
- 74. Proportions 1 2 linear ratio 1 : 2 area–ratio 1 : 4 = 22 linear ratio 1 : r area–ratio 1 : r2 1 r : : In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the area–measurements is 1 : (2)2 or 1 : 4 ratio. So it would take 4 times as much paint to paint the larger one. Given two similar objects with linear ratio 1 : r then their area–ratio is 1 : r2. This is also true for the surface areas.
- 75. Proportions 1 2 linear ratio 1 : 2 area–ratio 1 : 4 = 22 linear ratio 1 : r area–ratio 1 : r2 1 r : : In the definition of similarity, the measurements in proportion must be linear. It's not true that the ratio of areas is proportional to linear ratio of two similar objects. Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the area–measurements is 1 : (2)2 or 1 : 4 ratio. So it would take 4 times as much paint to paint the larger one. Given two similar objects with linear ratio 1 : r then their area–ratio is 1 : r2. This is also true for the surface areas i.e. two similar 3D solids of 1 : r linear ratio have 1 : r2 as their surface–area ratio.
- 76. Proportions The volume of an object is the measurement of the 3-dimensional space the object occupies.
- 77. Proportions Because the ratio of the linear measurements is 1 : 2, : linear ratio 1 : 2 1 2 The volume of an object is the measurement of the 3-dimensional space the object occupies.
- 78. Proportions Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the volume–measurements is 1 : (2)3 or 1 : 8 ratio. : linear ratio 1 : 2 volume–ratio 1 : 8 = 23 1 2 The volume of an object is the measurement of the 3-dimensional space the object occupies.
- 79. Proportions Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the volume–measurements is 1 : (2)3 or 1 : 8 ratio. So the weight of large one is 8 times as much as the small one (if they are made of the same thing).* : linear ratio 1 : 2 volume–ratio 1 : 8 = 23 1 2 The volume of an object is the measurement of the 3-dimensional space the object occupies.
- 80. Proportions Because the ratio of the linear measurements is 1 : 2, we may conclude that the ratio of the volume–measurements is 1 : (2)3 or 1 : 8 ratio. So the weight of large one is 8 times as much as the small one (if they are made of the same thing).* Given two similar objects with linear ratio 1 : r then their volume–ratio is 1 : r3. linear ratio 1 : r volume–ratio 1 : r3 :: linear ratio 1 : 2 volume–ratio 1 : 8 = 23 1 2 1 r The volume of an object is the measurement of the 3-dimensional space the object occupies.
- 81. The Golden Ratio Proportions
- 82. The Golden Ratio Proportions An important ratio in arts and science is the golden ratio.
- 83. The Golden Ratio Proportions An important ratio in arts and science is the golden ratio. Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”,
- 84. The Golden Ratio Proportions An important ratio in arts and science is the golden ratio. Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1
- 85. The Golden Ratio Proportions Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L An important ratio in arts and science is the golden ratio.
- 86. The Golden Ratio Proportions Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L An important ratio in arts and science is the golden ratio. 1 L
- 87. The Golden Ratio Proportions Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L =
- 88. The Golden Ratio Proportions Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L = The ratio1/L is the Golden Ratio.
- 89. The Golden Ratio Proportions L (1 – L) 1 L Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L = Since s = 1 – L, we’ve = The ratio1/L is the Golden Ratio.
- 90. The Golden Ratio Proportions L (1 – L) 1 L cross multiply Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L = Since s = 1 – L, we’ve = 1 – L = L2 The ratio1/L is the Golden Ratio.
- 91. The Golden Ratio Proportions L (1 – L) 1 L cross multiply 0 = L2 + L – 1 Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L = Since s = 1 – L, we’ve = 1 – L = L2 The ratio1/L is the Golden Ratio.
- 92. The Golden Ratio Proportions L (1 – L) 1 L cross multiply 0 = L2 + L – 1 Cut a line segment of length 1 into two, a long one L and a short one s such that the ratio of “1 to L” is the same as “L to s”, 1 L s An important ratio in arts and science is the golden ratio. L s 1 L = Since s = 1 – L, we’ve = 1 – L = L2 We’ll see later that the answer for L is about 0.618. http://en.wikipedia.org/wiki/Golden_ratio http://en.wikipedia.org/wiki/Golden_spiral The ratio1/L is the Golden Ratio.
- 93. Ex. A. Solve for x. Proportions 3 2 2x + 1 3 = 21 x 3 =1. –2x/7 –5 1½ 3 =9. 2. 6 2=3. x + 1 –57 =4. –2x + 1 6 =5. x + 1 23x =6. 5x + 1 3 4x = 8 2/3 7/5 x 14 =7. 9 4/3 3 2x/5 =8. 10. – x + 2 2 –4 3 =11. 3 2x + 1 1 x – 2 =12. x – 5 2 2x – 3 3 =13. 3 2 x – 4 x – 1 =14. –3 5 2x + 1 3x =15. 3x + 2 2 2x + 1 3 =16. 3 2 2x + 1 x – 3 =17. 2 x x + 1 3 =18. x 2 x + 4 x =19. 1 2
- 94. Ex. B. (Solve each problem. It’s easier if fractional proportions are rewritten as proportions of integers.) Proportions 20. If 4 cups of flour need 3 tsp of salt, then 10 cups flour need how much of salt? Different cookie recipes are given, find the missing amounts. 21. If 5 cups of flour need 7 cups of water then 10 cups water need how much of flour? 23. If 2 ½ tsp of butter flour need 3/4 tsp of salt, then 6 tsp of salt need how much of flour. 22. If 5 cups of flour need 7 cups of water then 10 cups flour need how much of water? 24. If 1¼ inches equals 5 miles real distance, how many miles is 5 inches on the map? For the given map scales below, find the missing amounts. 25. If 2½ inches equals 140 miles real distance, how many inches on the map correspond to the distance of 1,100 miles?
- 95. Proportions A. B. C. 3 4 1 5 5 6 26. x 20 x 20 6 27. x 3x + 4 2xx+2 15 x x 2x – 3 28. 29. 30. 31. 32–37. Find the surface areas of each cat above if A’s surface area is 2 ft2, B’s surfaces area is 7 ft2. and C’s is 280 ft2? Ex. C. 26 – 31. Solve for x. Use the given A, B and C,. 38–43. Find the weight of each cat above if cat A is 20 lbs, cat B is 8 lb. and cat C is 350 lb?
- 96. Rational Equations Word-Problems
- 97. Rational Equations Word-Problems We look at the following applications of rational equations.
- 98. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations.
- 99. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = .C B
- 100. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations.C B
- 101. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. The calculation of “per unit” is a good example:. C B
- 102. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units The calculation of “per unit” is a good example: Per unit amount = C B
- 103. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B
- 104. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B Total cost Number of units Per unit cost =
- 105. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units For example, a group of 5 people rent a taxi that cost $20 so each person’s share is 20/5 or $4 per person. Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B Total cost Number of units Per unit cost =
- 106. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units For example, a group of 5 people rent a taxi that cost $20 so each person’s share is 20/5 or $4 per person. However if one person backs out, then the cost goes up to s = 20/4 = $5 per person. Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B Total cost Number of units Per unit cost =
- 107. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units For example, a group of 5 people rent a taxi that cost $20 so each person’s share is 20/5 or $4 per person. However if one person backs out, then the cost goes up to s = 20/4 = $5 per person. Each remaining person has to pay $1 more. Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B Total cost Number of units Per unit cost =
- 108. Rational Equations Word-Problems Problems from the Multiplication–Division Operations We look at the following applications of rational equations. All the multiplicative formulas of the form AB = C may be written as A = . This divisional form leads to rational equations. Total amount Number of units For example, a group of 5 people rent a taxi that cost $20 so each person’s share is 20/5 or $4 per person. However if one person backs out, then the cost goes up to s = 20/4 = $5 per person. Each remaining person has to pay $1 more. Let’s formulate this as a word problem and solve it using rational equations. Cost per Unit The calculation of “per unit” is a good example: Per unit amount = C B Total cost Number of units Per unit cost =
- 109. Rational Equations Word-Problems A table is useful in organizing repeated calculations for comparing the results.
- 110. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results.
- 111. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 20 Total Cost No. of people
- 112. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people
- 113. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1)
- 114. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) Let’s compare the two different per/person costs. 20 (x – 1) 20 x
- 115. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) Let’s compare the two different per/person costs. cost more cost less 20 (x – 1) 20 x
- 116. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? A table is useful in organizing repeated calculations for comparing the results. Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) Let’s compare the two different per/person costs. cost more cost less ($1 more) 20 (x – 1) 20 x
- 117. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) 20 x Let’s compare the two different per/person costs. – = 1 cost more cost less ($1 more) A table is useful in organizing repeated calculations for comparing the results.
- 118. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) Let’s compare the two different per/person costs. – = 1[ ] x (x – 1) clear the denominators by LCM A table is useful in organizing repeated calculations for comparing the results. 20 x
- 119. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) 20 x Let’s compare the two different per/person costs. – = 1[ ] x (x – 1) clear the denominators by LCM x (x – 1) x (x – 1) A table is useful in organizing repeated calculations for comparing the results.
- 120. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) Let’s compare the two different per/person costs. = 1[ ] x (x – 1) clear the denominators by LCM x (x – 1) x (x – 1) 20x – 20(x – 1) = x(x – 1) A table is useful in organizing repeated calculations for comparing the results. 20 x –
- 121. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) Let’s compare the two different per/person costs. = 1[ ] x (x – 1) clear the denominators by LCM x (x – 1) x (x – 1) 20x – 20(x – 1) = x(x – 1) 20x – 20x + 20 = x2 – x A table is useful in organizing repeated calculations for comparing the results. 20 x –
- 122. Rational Equations Word-Problems Example A. A taxi is rented by a group of x people for $20 and the cost is shared equally. If there is one less person in the group then each of the remaining people has to pay $1 more. What is x? Total Cost No. of People Cost per Person = 20 x 20 (x – 1) Total Cost No. of people 20 x 20 (x – 1) 20 (x – 1) Let’s compare the two different per/person costs. = 1[ ] x (x – 1) clear the denominators by LCM x (x – 1) x (x – 1) 20x – 20(x – 1) = x(x – 1) A table is useful in organizing repeated calculations for comparing the results. 0 = x2 – x – 20 set one side as 0 20 x – 20x – 20x + 20 = x2 – x
- 123. Rational Equations Word-Problems 0 = x2 – x – 20
- 124. Rational Equations Word-Problems 0 = x2 – x – 20 0 = (x – 5)(x + 4)
- 125. Rational Equations Word-Problems 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4.
- 126. Rational Equations Word-Problems 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 127. Rational Equations Word-Problems Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 128. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 129. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 130. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Going back from B to A our rate was 2 mph so the return took 6/2 = 3 hours. Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 131. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Going back from B to A our rate was 2 mph so the return took 6/2 = 3 hours. In a table, Distance (m) Rate (mph) Time = D/R (hr) Going Return Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 132. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Going back from B to A our rate was 2 mph so the return took 6/2 = 3 hours. In a table, Distance (m) Rate (mph) Time = D/R (hr) Going 6 3 2 Return Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 133. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Going back from B to A our rate was 2 mph so the return took 6/2 = 3 hours. In a table, Distance (m) Rate (mph) Time = D/R (hr) Going 6 3 2 Return 6 2 3 Rate–Time–Distance 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Hence there are 5 people.
- 134. Rational Equations Word-Problems Let R = rate (mph), T = time (hours) and D = Distance (m) then RT = D or that T = . .D R We hiked a 6–mile trail from A to B at a rate of 3 mph so the trip took 6/3 = 2 hours. Going back from B to A our rate was 2 mph so the return took 6/2 = 3 hours. In a table, Distance (m) Rate (mph) Time = D/R (hr) Going 6 3 2 Return 6 2 3 Let’s turn this example into a rational–equation word problem. 0 = x2 – x – 20 0 = (x – 5)(x + 4) x = 5 or x = –4. Rate–Time–Distance Hence there are 5 people.
- 135. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x?
- 136. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? The rate of the return is “1 mph faster then x” so it’s (x + 1).
- 137. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go Return The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 138. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 Return 6 The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 139. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x Return 6 The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 140. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 141. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 142. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table
- 143. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table Now we use the information about the times.
- 144. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x number of hrs for going Now we use the information about the times. number of hrs for returning
- 145. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. (total trip 5 hrs)number of hrs for going number of hrs for returning
- 146. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 (total trip 5 hrs)number of hrs for going number of hrs for returning
- 147. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 use the cross multiplication method
- 148. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 6(x + 1) + 6x x(x + 1) = 5 use the cross multiplication method
- 149. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 6(x + 1) + 6x x(x + 1) = 5 5 1 =x(x + 1) 12x + 6 use the cross multiplication method
- 150. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 6(x + 1) + 6x x(x + 1) = 5 5 1 =x(x + 1) 12x + 6 Cross again. You finish it. use the cross multiplication method
- 151. Rational Equations Word-Problems Example B. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 5 hours for the entire round trip. What is x? D = Distance (m) R = Rate (mph) Time = D/R (hr) Go 6 x 6/x Return 6 x + 1 6/(x + 1) The rate of the return is “1 mph faster then x” so it’s (x + 1). Put these into the table 6 (x + 1) 6 x Now we use the information about the times. + = 5 6(x + 1) + 6x x(x + 1) = 5 5 1 =x(x + 1) 12x + 6 Cross again. You finish it. (Remember that x = 2 mph.) use the cross multiplication method
- 152. Rational Equations Word-Problems Job–Rates
- 153. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our pizza–eating rate =
- 154. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate =
- 155. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr)
- 156. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our pizza–eating rate =
- 157. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate =
- 158. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3
- 159. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3 These types of rates (in per unit of time) are called job–rates.
- 160. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3 These types of rates (in per unit of time) are called job–rates. We note the following about job–rates.
- 161. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3 These types of rates (in per unit of time) are called job–rates. We note the following about job–rates. * In all such problems, a complete job to be done may be viewed as a pizza needing to be eaten and it’s set to be 1.
- 162. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3 These types of rates (in per unit of time) are called job–rates. We note the following about job–rates. * In all such problems, a complete job to be done may be viewed as a pizza needing to be eaten and it’s set to be 1. Hence if it takes 3 hrs to paint a room, to mow a lawn, or to fill a pool, then the job–rate for each is 1 job 3 hours
- 163. Rational Equations Word-Problems Job–Rates If we can eat 6 pizzas in 2 hours then our 6 pizzas 2 hours pizza–eating rate = = 3 (piz/hr) If we can eat 2 pizzas in 6 hours then our 2 pizzas 6 hours pizza–eating rate = = (piz/hr)1 3 These types of rates (in per unit of time) are called job–rates. We note the following about job–rates. * In all such problems, a complete job to be done may be viewed as a pizza needing to be eaten and it’s set to be 1. Hence if it takes 3 hrs to paint a room, to mow a lawn, or to fill a pool, then the job–rate for each is 1 job 3 hours = (room), (lawn), or (pool) / hr1 3
- 164. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed.
- 165. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours
- 166. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = 2 3 1 6 *
- 167. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = 3
- 168. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. 3
- 169. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. 3
- 170. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. Example C. We can paint 3/4 of a wall in 4 ½ hrs, what is the job–rate? How long would it take to paint the entire wall? 3
- 171. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. Example C. We can paint 3/4 of a wall in 4 ½ hrs, what is the job–rate? How long would it take to paint the entire wall? The job–rate is 3/4 4½ 3
- 172. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. Example C. We can paint 3/4 of a wall in 4 ½ hrs, what is the job–rate? How long would it take to paint the entire wall? The job–rate is 3/4 4½ = 3/4 9/2 = 3 4 2 9 * 3
- 173. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. Example C. We can paint 3/4 of a wall in 4 ½ hrs, what is the job–rate? How long would it take to paint the entire wall? The job–rate is 3/4 4½ (wall / hr)= 1 6= 3/4 9/2 = 3 4 2 9 * 2 3 3
- 174. Rational Equations Word-Problems * The job–rate may be computed even if only a portion of the job is completed. For example, if it took us 6 hrs to paint 2/3 of the room then the job–rate is 2/3 room 6 hours = (room/hr)1 9 2 3 1 6 * = * The reciprocal of a job–rate is the amount of time it would take to complete one job. (lawn/hr), 1 9 Hence if the job–rate is 9 1 then it would take = 9 hrs to complete the entire lawn. Example C. We can paint 3/4 of a wall in 4 ½ hrs, what is the job–rate? How long would it take to paint the entire wall? The job–rate is 3/4 4½ (wall / hr)= 1 6= 3/4 9/2 = 3 4 2 9 * 2 3 The reciprocal of this job–rate is 6 (hr / wall) or it would take 6 hours to paint the entire wall. 3
- 175. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe?
- 176. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A B
- 177. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A B
- 178. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 B
- 179. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 The rate of A = tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B
- 180. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 The rate of A = tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 181. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 The rate of A = tank/hr. The rate of B = 3/4 1½ = Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 182. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 = The rate of A = tank/hr. The rate of B = 3/4 1½ = 3/4 3/2 Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 183. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 = 1 2 The rate of A = tank/hr. The rate of B = tank/hr. 3/4 1½ = 3/4 3/2 Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 184. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 = 1 2 The rate of A = tank/hr. The rate of B = tank/hr. 3/4 1½ = 3/4 3/2 b. How much time would it take for pipe B to fill the tank alone? Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 185. Rational Equations Word-Problems Example D. Pipe A can fill a full tank of water in 3 hours, pipe B can fill ¾ of the (same) tank of water in 1½ hr. a. What is unit of the fill–rate and what is the rate of each pipe? The unit of the rate in question is tank/hr. 1 3 = 1 2 The rate of A = tank/hr. The rate of B = tank/hr. 3/4 1½ = 3/4 3/2 b. How much time would it take for pipe B to fill the full tank alone? We reciprocate the rate of B “½ (tank/hr)” and get 2 hr/tank or that it would take 2hrs for pipe B to fill a full tank. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2
- 186. Rational Equations Word-Problems Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used?
- 187. Rational Equations Word-Problems The combined rate is the sum of the rates Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used? 1 3 + 1 2
- 188. Rational Equations Word-Problems 5 6= The combined rate is the sum of the rates tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used? 1 3 + 1 2
- 189. Rational Equations Word-Problems 5 6= The combined rate is the sum of the rates tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 The reciprocal of the combined rate is 6/5 therefore it would take 1 1/5 hr if both pipes are used. c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used? 1 3 + 1 2
- 190. Rational Equations Word-Problems 5 6= The combined rate is the sum of the rates tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 The reciprocal of the combined rate is 6/5 therefore it would take 1 1/5 hr if both pipes are used. c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used? 1 3 + 1 2 We note that in the above example, it takes pipe B one hour less to fill the tank then pipe A does and that together they got the job done in 6/5 hr.
- 191. Rational Equations Word-Problems 5 6= The combined rate is the sum of the rates tank/hr. Pipes A = Amount (tank) T = Time (hr) Rate = A/T (tank/hr) A 1 3 1/3 B 3/4 1 ½ = 3/2 1/2 The reciprocal of the combined rate is 6/5 therefore it would take 1 1/5 hr if both pipes are used. c. What is the combined rate if both pipes are used and how long would it take to fill the entire tank if both pipes are used? 1 3 + 1 2 We note that in the above example, it takes pipe B one hour less to fill the tank then pipe A does and that together they got the job done in 6/5 hr. Let’s treat this as a word problem.
- 192. Rational Equations Word-Problems Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?)
- 193. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?)
- 194. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1)
- 195. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. .
- 196. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. Put these into a table.
- 197. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A B A & B Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. Put these into a table.
- 198. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x B (x – 1) A & B 6/5 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. Put these into a table.
- 199. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. Put these into a table.
- 200. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. . Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x.
- 201. Rational Equations Word-Problems Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. . 5 6= 1 x + 1 (x – 1) Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x.
- 202. Rational Equations Word-Problems 5 6= 1 x + 1 (x – 1) clear the denominatorsx (x – 1)][ Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) It takes 6/5 hr to use both. . Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x.
- 203. Rational Equations Word-Problems 5 6= 1 x + 1 (x – 1) clear the denominators6x (x – 1) 6x ][ Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x. It takes 6/5 hr to use both. . 6(x – 1) x (x – 1)
- 204. Rational Equations Word-Problems 5 6= 1 x + 1 (x – 1) clear the denominators6x (x – 1) 6x ][ Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x. It takes 6/5 hr to use both. . 6(x – 1) x (x – 1) 6x – 6 + 6x = 5x2 – 5x
- 205. Rational Equations Word-Problems 5 6= 1 x + 1 (x – 1) clear the denominators6x (x – 1) 6x ][ Set x = number of hours for A to fill the tank Pipes Time (hr) Rate (tank/hr) A x 1/x B (x – 1) 1/(x – 1) A & B 6/5 5/6 Example E. Pipe B takes one hour less to fill a full tank of water then pipe A does and together they got the job done in 6/5 hr. How long will it take pipe A to fill the tank alone? (What should x be?) So the number of hours for B to fill the tank is (x – 1) Put these into a table. The rates for A and B add to the combined rate 5/6 so we get a rational equation in x. It takes 6/5 hr to use both. . 6(x – 1) x (x – 1) 6x – 6 + 6x = 5x2 – 5x 0 = 5x2 – 17x + 6 (Question: Why is the other solution no good?) you finish this...
- 206. Rational Equations Word-Problems Ex. For problems 1 – 9, fill in the table, set up an equation and solve for x. (If you can guess the answer first, go ahead. But set up and solve the problem algebraically to confirm it.) 2. A group of x people are to share 6 slices of pizza equally. If there are three less people in the group then each person would get one more slice. What is x? 3. A group of x people are to share 6 slices of pizza equally. If one more person joins in to share the pizza then each person would get three less slices. What is x? 1. A group of x people are to share 6 slices of pizza equally. If there is one less person in the group then each person would get one more slice. What is x? Total No. of People Per Person = Total No. of people
- 207. Rational Equations Word-Problems 5. A group of x people are to share 12 slices of pizza equally. If two more people want to partake then each person would get one less slice. What is x? 6. A group of x people are to share 12 slices of pizza equally. If four people in the group failed to show up then each person would get four more slices. What is x? 7. A group of x people are to share the cost of $24 to rent a van equally. If three more people want to partake then each person would pay $4 less. What is x? 8. A group of x people are to share the cost of $24 to rent a van equally. If four more people join in the group then each person would pay $1 less. What is x? 9. A group of x people are to share the cost of $24 to rent a van equally. If there are two less people in the group then each person would pay $1 more. What is x?
- 208. Rational Equations Word-Problems Distance (m) Rate (mph) Time = D/R (hr) 1st trip 2nd trip 10. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph slower. It took us 5 hours for the entire round trip. What is x? HW C. For problems 10 – 16, fill in the table, set up an equation and solve for x. 11. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 3 mph faster. It took us 3 hours for the entire round trip. What is x? 12. We hiked 6 miles from A to B at a rate of x mph. For the return trip our rate was 1 mph faster. It took us 9 hours for the entire round trip. What is x?
- 209. Rational Equations Word-Problems 13. We hiked 3 miles from A to B at a rate of x mph. Then we hiked 2 miles from B to C at a rate that was 1 mph slower. It took us 2 hours for the entire round trip. What is x? 14. We hiked 6 miles from A to B at a rate of x mph. Then we hiked 3 miles from B to C at a rate that was 1 mph faster. It took us 3 hours for the entire round trip. What is x? 15. We ran 8 miles from A to B. Then we ran 5 miles from B to C at a rate that was 1 mph faster. It took one more hour to get from A to B than from B to C. What was the rate we ran from A to B? 16. We piloted a boat 12 miles upstream from A to B. Then we piloted it downstream from B back to A at a rate that was 4 mph faster. It took one more hour to go upstream from A to B than going downstream from B to C. What was the rate of the boat going up stream from A to B?
- 210. Rational Equations Word-Problems 17. Its takes 3 hrs to complete 2/3 of a job. What is the job–rate? How long would it take to complete the whole job? How long would it take to complete ½ of the job? For problems 17–22, don’t set up equations. Find the answers directly and leave the answer in fractional hours. 18. Its takes 4 hrs to complete 3/5 of a job. What is the job–rate? How long would it take to complete the whole job? How long would it take to complete 1/3 of the job? Time Rate A B A & B 19. Its takes 4 hrs for A to complete 3/5 of a job and 3 hr for B to do 2/3 of the job. Complete the table. What is their combined job–rate? How long would it take for A & B to complete the whole job together?
- 211. Rational Equations Word-Problems Time Rate A B A & B 20. Its takes 1/2 hr for A to complete 2/5 of a job and 3/4 hr for B to do 1/6 of the job. Complete the table. What is their combined job–rate? How long would it take for A & B to complete the whole job together? Time Rate A B A & B 21. It takes 3 hrs for pipe A to fill a pool and 5 hrs for pipe B to drain the pool. Complete the table. If we use both pipes to fill and drain simultaneously, what is their combined job–rate? Will the pool be filled eventually? If so, how long would it take for the pool to fill?
- 212. Rational Equations Word-Problems Time Rate leak pump both 22. A leak in our boat will fill the boat in 6 hrs at which time the boat sinks. We have a pump that can empty a filled boat in 9 hrs. How much time do we have before the boat sinks? For 23 –26, given the information, how many hours it would take for A or B to do the job alone. 23. It takes A one more hour than B to do the job alone. It takes them 2/3 hr to do the job together. 24. It takes A two more hours than B to do the job alone. It takes them 3/4 hr to do the job together. 25. It takes B two hour less than A to do the job alone. It takes them 4/3 hr to do the job together. 26. It takes A three more hours than B to do the job alone. It takes them 2 hr to do the job together.

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