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Inverse Variation

Inverse Variation. ALGEBRA 1 LESSON 8-10. (For help, go to Lesson 5-5.). Suppose y varies directly with x. Find each constant of variation. 1. y = 5 x 2. y = –7 x 3. 3 y = x 4. 0.25 y = x Write an equation of the direct variation that includes the given point.

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Inverse Variation

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  1. Inverse Variation ALGEBRA 1 LESSON 8-10 (For help, go to Lesson 5-5.) Suppose y varies directly with x. Find each constant of variation. 1.y = 5x2.y = –7x3. 3y = x4. 0.25y = x Write an equation of the direct variation that includes the given point. 5. (2, 4) 6. (3, 1.5) 7. (–4, 1) 8. (–5, –2) 8-10

  2. Inverse Variation ALGEBRA 1 LESSON 8-10 1.y = 5x; constant of variation = 5 2. y = –7x; constant of variation = –7 3. 3y = x • 3y = • x y = x; constant of variation = 4. 0.25y = x y = x 4 • y = 4 • x y = 4x; constant of variation = 4 Solutions 1 3 1 3 1 3 1 3 1 4 1 4 8-10

  3. Inverse Variation ALGEBRA 1 LESSON 8-10 Solutions (continued) 5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x. 6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x. 7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – and y = – x. 8. Point (–5, –2) in y = kx: –2 = k(–5), so k = and y = x. 1 4 1 4 2 5 2 5 8-10

  4. 72 x The equation of the inverse variation is xy = 72 or y = . Inverse Variation ALGEBRA 1 LESSON 8-10 Suppose y varies inversely with x, and y = 9 when x = 8. Write an equation for the inverse variation. xy = kUse the general form for an inverse variation. (8)(9) = kSubstitute 8 for x and 9 for y. 72 = kMultiply to solve for k. xy = 72 Write an equation. Substitute 72 for k in xy = k. 8-10

  5. Inverse Variation ALGEBRA 1 LESSON 8-10 The points (5, 6) and (3, y) are two points on the graph of an inverse variation. Find the missing value. x1 • y1 = x2 • y2Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation. 5(6) = 3y2Substitute 5 for x1, 6 for y1, and 3 for x2. 30 = 3y2Simplify. 10 = y2Solve for y2. The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6). 8-10

  6. Relate:  A weight of 130 lb is 5 ft from the fulcrum.A weight of 93 lb is x ft from the fulcrum.Weight and distance vary inversely. Define:  Let weight1 = 130 lbLet weight2 = 93 lbLet distance1 = 5 ftLet distance2 = x ft Inverse Variation ALGEBRA 1 LESSON 8-10 Jeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If Tracy weighs 93 pounds, how far from the fulcrum should she sit in order to balance the lever? 8-10

  7. Write: weight1 • distance1 = weight2 • distance2 130 • 5 = 93 • xSubstitute. 650 93 = xSolve for x. 6.99 = xSimplify. Inverse Variation ALGEBRA 1 LESSON 8-10 (continued) 650 = 93xSimplify. Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever. 8-10

  8. a. xy 3 10 5 6 10 3 Check each product xy. xy: 3(10) = 30    5(6) = 30    10(3) = 30 Inverse Variation ALGEBRA 1 LESSON 8-10 Decide if each data set represents a direct variation or an inverse variation. Then write an equation to model the data. The values of y seem to vary inversely with the values of x. The product of xy is the same for all pairs of data. So, this is an inverse variation, and k = 30. The equation is xy = 30. 8-10

  9. y x b. xy 2 3 4 6 8 12 Check each ratio . 12 8 6 4 = 1.5 y x 3 2 = 1.5 = 1.5 y x The ratio is the same for all pairs of data. Inverse Variation ALGEBRA 1 LESSON 8-10 (continued) The values of y seem to vary directly with the values of x. So, this is a direct variation, and k = 1.5. The equation is y = 1.5x. 8-10

  10. cost souvenirs Since the ratio is constant at $10 each, this is a direct variation. Inverse Variation ALGEBRA 1 LESSON 8-10 Explain whether each situation represents a direct variation or an inverse variation. a. You buy several souvenirs for $10 each. The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs. b. The cost of a $25 birthday present is split among several friends. The cost per person times the number of people equals the total cost of the gift. Since the total cost is a constant product of $25, this is an inverse variation. 8-10

  11. xy 1 2 6 1 3 1 6 1 3 xy = 1 9 3 1 18 Inverse Variation ALGEBRA 1 LESSON 8-10 1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y. 2. Find the constant of variation k for the inverse variation where a = 2.5 when b = 7. 0.5 17.5 3. Write an equation to model the data and complete the table. 4. Tell whether each situation represents a direct variation or an inverse variation. a. You buy several notebooks for $3 each. b. The $45 cost of a dinner at a restaurant is split among several people. direct variation Inverse variation 8-10

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