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| Question 206918:  This is not for homework; however, it is one in a homework text for the University of Phoenix Introductory and Intermediate Algebra.
 In order to disuss with my classmates about how to solve linear equations, I need to be able to understand this one by today.  Rod is a pilot for Crossland Airways. He computes his flight time against a headwind for a tip of 2900 mi. at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great.  Find the headwind and the plane's air speed.
 I understand I must use d=rxt. How do I come up with the two equations?  I thought maybe it was 2,900= R x 5  and D = r 1/2 x 4.50; but, they do not look right.  I came up with an air speed of 124.7?  Please help.Reference:
 Bittenger, M. L. & Beecher, J. A. (2007). Introductory and intermediate algebra (3rd ed.). Boston: Pearson-Addison Wesley.
 
 Found 2 solutions by  Alan3354, Theo:
 Answer by Alan3354(69443)
      (Show Source): 
You can put this solution on YOUR website! Rod is a pilot for Crossland Airways. He computes his flight time against a headwind for a tip of 2900 mi. at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great. Find the headwind and the plane's air speed. ------------------
 a = airspeed
 w = wind speed
 2900 = (a-w)*5
 2900 = (a-w/2)*(29/6)
 ------------
 a- w = 580
 a- w/2 = 600
 ------------ Subtract
 -w/2 = -20
 w = 40
 --------
 a = 620
Answer by Theo(13342)
      (Show Source): 
You can put this solution on YOUR website! Rod is a pilot for Crossland Airways. He computes his flight time against a headwind for a trip of 2900 mi. at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great.  Find the headwind and the plane's air speed. -----
 let x = rate of speed of airplane.
 let y = rate of speed of headwind
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 let D = distance = 2900 miles
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 let T = time in minutes
 5 hours = 5 hours = 300 minutes
 4 hours 50 minutes = 290 minutes
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 converting to minutes should make the arithmetic a little easier even though the number will get large.
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 formula is R * T = D where R is the combined rate of speed and T is the time it takes in minutes and D is the distance in miles.
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 let R = (x-y)
 let T = 300 minutes
 formula becomes:
 (x - y) * 300 = 2900
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 if the rate of the headwind is half as great, then the rate of the headwind becomes y/2
 -----
 let R = (x - (y/2))
 Let T = 290 minutes
 formula becomes:
 (x - (y/2)) * 290 = 2900
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 we have 2 formulas to solve simultaneously.
 (x - y) * 300 = 2900
 (x - (y/2)) * 290 = 2900
 -----
 remove parentheses and simplify to get:
 300*x - 300*y = 2900
 290*x - 145*y = 2900
 -----
 you can solve these in several ways.
 -----
 solve by substitution means you use one of the equations to get x in terms of y or y in terms of x and then substitute in the other equation.
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 solve by elimination means you multiply both sides of the equations by factors that will allow one of the unknown variables to cancel out once you add the equations together.
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 since both equations equal the same value, you can set each equation equal to each other.
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 we'll solve by elimination.
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 your equations after simplification are:
 300*x - 300*y = 2900
 290*x - 145*y = 2900
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 if you multiply the first equation by 290 and you multiply the second equation by 300 then you should be able to eliminate the x when you add or subtract both equations from each other.
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 after multiplication, the equations become:
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 87000*x - 87000*y = 841000
 87000*x - 43500*y = 870000
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 subtract the second equation from the first equation to get:
 - 43500 * y = - 29000
 divide both sides of this equation by (-43500) to get y = .667 = 2/3 mile per minute.
 if y = 2/3, then y/2 = 1/3 mile per minute.
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 now that we know the rate of speed of the headwind, we can solve for the rate of speed of the airplane
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 in our first equation, we had:
 300*x - 300*y = 2900
 this equation becomes:
 300*x - 300*(2/3) = 2900 which becomes:
 300*x - 200 = 2900 which becomes:
 300*x = 3100 which becomes:
 x = 10 and 1/3 equals 31/3 miles per minute.
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 in our second equation, we now have:
 290*(31/3) - 290*(1/3) = 2900 which becomes:
 8990/3 - 290/3 = 2900 which becomes:
 8700/3 = 2900 which becomes:
 2900 = 2900 which means that the value of 31/3 miles per minute for the plane and the value of 2/3 miles per minute for the headwind is accurate.
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 converting these back to hours we get:
 speed of the plane is 60*31/3 = 620 miles per hour.
 speed of the headwind is 60*2/3 = 40 miles per hour.
 -----
 you can plug these values into the original equations to see that they work as well.
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