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Chapter 6 Section 7 Rational Equations: Applications and Problem Solving Setup and Solve Applications Containing Rational Equations: Steps. 1. Read 2. Reread – to make sure that you understand the problem. Look for key words. 3. Write the equation 4. Solve 5. Check 6. Answer the question originally asked. Rational Equations: Applications and Problem Solving We need to check our solutions to make sure. 1. We did not make a mistake in our calculations. 2. The solution is valid for the original equation. 3. The solution is correct for the content. Example: a negative number would not be correct for the length or width. Types of Application Problems Geometry 1. Area of Rectangle = length ● width A = LW 2. Area of Triangle = ½ base ● height A = ½bh 3. Area of a Square = side2 A = s2 Types of Application Problems Motion 1. distance = rate ● time d = rt 2. time = distance / rate t = d/r 3. rate = time / distance r= t/d Types of Application Problems Work 1. Amount = rate ● time A = rt Number 1. Two numbers are Reciprocals of each other when their product is 1. example: the reciprocal of 3 is 1/3 the reciprocal of 3/5 is 5/3 0 has no reciprocal Geometry Problem - The area of a rectangle is 160 square meters. Determine the length and width if the width is 6 meters less than the length. A = LW Let x = length X = L = 16 W = (x – 6) = 16 – 6 = 10 Then (x - 6)(x) = 160 x2 – 6x = 160 x2 – 6x – 160 = 0 (x – 16 )(x + 10) = 0 x - 16 = 0 x + 10 = 0 x = 16 x ≠ -10 -10 not an answer Number Problem - One number is 5 times another. The difference of their reciprocals is 2/5. Determine the numbers. Let x = first number 1 reciprocal = x 5x = second number reciprocal = 1 1 2 x 5x 5 5 1 2x 1st number = x = 2 LCD 5 x 2 1 1 (5 x) (5 x) 5 x 5x 4 2x 1 5x 2nd number = 5x = (5)(2) = 10 1 1 2 (5 x ) ( 5x ) ( 5 x ) x 5x 5 2 x Motion Problem - The current in the river is 3 mph. If it takes the same amount of time to travel 14 miles downstream as 2 miles upstream, find the speed of the boat in still water. r = still water Direction r + 3 = downstream (with current) Downstream r – 3 = upstream (against current) Upstream d t r Distance Rate Time 14 r+3 14/(r+3) 2 r-3 2/(r-3) Because the time it takes to travel upstream is the same as downstream set times equal to each other and cross multiply to solve. 14 2 r 3 r 3 14r 42 2r 6 (14)(r 3) (2)(r 3) 14r 2r 6 42 (14r 42) (2r 6) 12r 48 The speed of the boat in still water is 4 mph r4 Motion Problem - A person rides a bike on a trail part of the time at 6 mph and part of the time at 10 mph. If a total distance of 28 miles is traveled in 4 hours, how long did the person ride at each speed? d = at 10 mph t 28 - d = at 6 mph d r Direction Distance Rate Time A d = 10 10 d/10=10/10 1 hour at 10 mph B 28-d = 18 6 (28-d)/6 = (28-10)/6 = 18/6 3 hours at 6 mph d 28 d 4 10 6 6d 10d 240 280 6 d 28 d 10 60 60 (4)(60) 10 6 4d 40 d 10 6d (280 10d ) 240 Motion Problem - Tom runs 8 mph and Jake runs at 6 mph If they start at the same time and Tom finishes the course 1.5 hours ahead of Jake, Determine the length of the course. d = length of course d t r Direction Distance Rate Time Tom d 8 d/8 = 36/8 = 4.5 Jake d 6 d/6 = 36/6 = 6 d d -1.5 8 6 6d 72 8d 6 d d 8 48 48 1.5(48) 8 6 72 8d 6d 72 2d Distance traveled is 36 miles. Check: 6 - 4.5 = 1.5 hours ahead of Jake. 6d 8d 72 36 d Work Problem – One person can do a task in 6 hours. A second person can do a task in 4 hours. How long will it take them working together to do the task? a = rt t = time working together Rate of Work Time Worked Part of Task 1st person 6 hrs t t/6 2nd person 4 hrs t t/4 t t =1 6 4 4t 6t 24 4 t t 6 24 24 (1)(24) 6 4 10t 24 t 24 /10 t 2.4 It takes 2.4 hours to complete the task working together. Work Problem – One pipe can fill a tank in 4 hours. Another pipe can empty a tank in 6 hours. If both pips are open how long will it take to fill the empty tank. t = time to fill the tank both valves open Rate of Work Time Part of Task 1st tank 1/6 hrs t t/6 2nd tank 1/4 hrs t t/4 t t 1 6 4 4t 6t 24 4 t t 6 24 24 (1)(24) 6 4 -2t 24 t 24 / 2 It takes 12 hours to fill the tank with both valves open. t 12 Work Problem – Tom and Frank can paint an apartment in 12 hours. Tom can paint the apartment by himself in 18 hours. How long does it take Frank to paint the apartment by himself? t = time Frank to paint apartment by himself Rate of Work Time Part of Task Tom 1/18 12 12/18 Frank 1/t 12 12/t 12 12 1 18 t 12t 216 18t 12 18 t 18 12 18 t (1)(18t ) t 216 18t 12t 216 6t 36 t It takes Frank 36 hours to paint the apartment by himself. Work Problem – Bill by himself can sweep up sand in a parking lot in 20 hours. Sally by herself can sweep up the sand in 25 hours. After Sally sweeps sand for 10 hours, she stops and Bill takes over. How long will it take Bill to finish the job? t = time Frank to paint apartment by himself Rate of Work Time Part of Task Bill 1/20 t t/20 Sally 1/25 10 10/25 = 2/5 2 t 1 5 20 8 t 20 4 2 t 20 20 (1)(20) 5 20 t 20 8 t 12 It takes Bill 12 hours to finish the job by himself. Remember It is important that you understanding the problem, rather than memorizing the procedures. Work Problems: times for working together will be less than times for working alone. Rectangle: area = (length)(width) distance = (rate)(time) It is important that you check the answer in the equation as well as making sure your answer is reasonable If the equation is setup wrong the equation could be right but the solution could still be incorrect HOMEWORK 6.7 Page 405: #5, 11, 15, 17