Direct Variation Solve each equation for the given variable. ALGEBRA 1 LESSON 8-9 (For help, go to Lessons 2–7 and 4–1.) Solve each equation for the given variable. 1. nq = m; q 2. d = rt; r 3. ax + by = 0; y Solve each proportion. 4. = 5. = 6. = 7. = 8. = 9. = 5 8 x 12 4 9 n 45 25 15 y 3 7 n 35 50 8 d 20 36 14 18 63 n 8-9

Direct Variation Solutions 1. nq = m 2. d = rt 3. ax + by = 0 = = ALGEBRA 1 LESSON 8-9 Solutions 1. nq = m 2. d = rt = = q = = r r = nq n m d t rt 3. ax + by = 0 ax – ax + by = 0 – ax by = –ax = y = – by b –ax ax 4. = 8x = 5(12) 8x = 60 x = 7.5 5 8 x 12 5. = 6. = 9n = 4(45) 15y = 25(3) 9n = 180 15y = 75 n = 20 y = 5 4 9 n 45 25 15 y 3 8-9

Direct Variation Solutions (continued) 7. = 8. = 9. = ALGEBRA 1 LESSON 8-9 Solutions (continued) 7. = 8. = 35n = 7(50) 20d = 8(36) 35n = 350 20d = 288 n = 10 d = 14.4 7 n 35 50 8 d 20 36 14 18 63 9. = 14n = 18(63) 14n = 1134 n = 81 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 Is each equation a direct variation? If it is, find the constant of variation. a. 2x – 3y = 1 –3y = 1 – 2x Subtract 2x from each side. y = – + x Divide each side by –3. 1 3 2 The equation does not have the form y = kx. It is not a direct variation. b. 2x – 3y = 0 –3y = –2x Subtract 2x from each side. y = x Divide each side by –3. 2 3 The equation has the form y = kx, so the equation is a direct variation. The constant of variation is . 2 3 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 Write an equation for the direct variation that includes the point (–3, 2). y = kx Use the general form of a direct variation. 2 = k(–3) Substitute –3 for x and 2 for y. – = k Divide each side by –3 to solve for k. 2 3 y = – x Write an equation. Substitute – for k in y = kx. 2 3 The equation of the direct variation is y = – x . 2 3 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 The weight an object exerts on a scale varies directly with the mass of the object. If a bowling ball has a mass of 6 kg, the scale reads 59. Write an equation for the relationship between weight and mass. Relate: The weight varies directly with the mass. When x = 6, y = 59. Define: Let = the mass of an object. Let = the weight of an object. x y 8-9

Direct Variation (continued) ALGEBRA 1 LESSON 8-9 (continued) Write: = k Use the general form of a direct variation. 59 = k(6) Solve for k. Substitute 6 for x and 59 for y. x y = k Divide each side by 6 to solve for k. 59 6 y = x Write an equation. Substitute for k in y = kx. 59 6 The equation y = x relates the weight of an object to its mass. 59 6 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 For the data in each table, tell whether y varies directly with x. If it does, write an equation for the direct variation. 1 –2 = –0.5 –1 2 4 y x x y –2 1 2 –1 4 –2 2 –1 = –2 1 –4 = 2 y x x y –1 2 1 2 2 –4 No, the ratio is not the same for each pair of data. x y Yes, the constant of variation is –0.5. The equation is y = –0.5x. 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 Suppose a windlass requires 0.75 lb of force to lift an object that weighs 48 lb. How much force would you need to lift 210 lb? Relate: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb. Define: Let n = the force you need to lift 210 lb. Write: = Use a proportion. force1 weight1 force2 weight2 Substitute 0.75 for force1, 48 for weight1, and 210 for weight2. 0.75 48 n 210 = 0.75(210) = 48n Use cross products. Solve for n. n 3.3 You need about 3.3 lb of force to lift 210 lb. 8-9

Direct Variation ALGEBRA 1 LESSON 8-9 1. Is each equation a direct variation? If it is, find the constant of variation. a. x + 5y = 10 b. 3y + 8x = 0 no yes; – 8 3 2. Write an equation of the direct variation that includes the point (–5, –4). y = x 4 5 3. For each table, tell whether y varies directly with x. If it does, write an equation for the direct variation. x y –1 3 0 0 2 –6 3 –9 a. x y –1 –2 0 0 1 2 3 –6 b. yes; y = –3x no 8-9