Rational Functions: Applications Presented by: 1___________ 2___________

1) The difference of a whole number and its reciprocal is . 63 8 Number Problems 1) The difference of a whole number and its reciprocal is . 63 8 What is the number? a) Represent c) Solve Let x = the whole number x - = 1 x 63 8 8x 8x = reciprocal 1 x 8x2 – 8 = 63x 8x2 - 63x – 8 = 0 b) Equate (8x + 1) (x – 8) = 0 x - = 1 x 63 8 8x + 1 = 0 x – 8 = 0 8x = -1 x = 8 x = -1 8 Copyright © by Mr. Florben G. Mendoza

Work Problems 1) Two pipes are connected to the same tank. When working together, they can fill the tank in 2 hours. The larger pipe, working alone, can fill the tank in 3 hours less time than the smaller one. How long would the smaller one take, working alone, to fill the tank? b) Equate/Solve a) Represent + 2 t t - 3 = 1 t (t - 3) Smaller Pipe Larger Complete Job Together 2 1 Alone t t - 3 2t - 6 + 2t = t2 – 3t t2 + - 2t – 2t – 3t + 6 = 0 t2 - 7t + 6= 0 (t - 6) (t - 1) = 0 t - 6 = 0 t - 1= 0 t = 6 t = 1 Copyright © by Mr. Florben G. Mendoza

Work Problems: 1) Two pipes are connected to the same tank. When working together, they can fill the tank in 2 hours. The larger pipe, working alone, can fill the tank in 3 hours less time than the smaller one. How long would the smaller one take, working alone, to fill the tank? b) Equate/Solve a) Represent t = 6 t = 1 Smaller Pipe Larger Complete Job Together 2 1 Alone t t - 3 c) Substitute Smaller Pipe: 6 hours Larger Pipe: 3 hours Copyright © by Mr. Florben G. Mendoza

Distance Problems: 1) A truck traveled the first 100 kilometers of a trip at one speed and the last 135 kilometers at an average speed of 5 kilometers per hour less. If the entire trip took 5 hours, what was the average speed for the first part of the trip? 135 = 100 t - 5 (5 – t) b) Equate/Solve Solution: a) Represent d = rt d r t 100 135 r - 5 5 - t 100 – 5t t 135 = (5 – t) t t 135t = (100 – 5t) (5 – t) t2 – 52t + 100 = 0 135t = 500 – 100t – 25t + 5t2 (t – 50) (t – 2) = 0 100 = r t r = 100/t 5t2 – 260t + 500 = 0 t – 50 = 0 t – 2 = 0 135 = (r – 5) (5 – t) t2 – 52t + 100 = 0 t = 50 t = 2 t2 – 52t + 100 = 0 Copyright © by Mr. Florben G. Mendoza

Distance Problems: 2) A truck traveled the first 100 kilometers of a trip at one speed and the last 135 kilometers at an average speed of 5 kilometers per hour less. If the entire trip took 5 hours, what was the average speed for the first part of the trip? t = 50 Solution: b) t = 2 a) Let d r t 100 r = d/t r = 100/2 r = 50 kph 2 hrs 135 r – 5 50 – 5 = r = 45 kph 5 – t 5 – 2 = 3hrs c) d r t 100 135 r - 5 5 - t d = rt 100 = r t r = 100/t 135 = (r – 5) (5 – t) Copyright © by Mr. Florben G. Mendoza

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