Question 164682
D = 780 miles
let A = airspeed
let W = windspeed
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since it takes more time to go west than east, it's fair to say that the wind was against the airplane going west and with the airplane going east.
TW = time going west = 2 hours
TE = time going east = 1.5 hours
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since the rate of travel times the duration of travel equals the distance, the equations for going west and going east are:
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TW * (A-W) = 780 
TE * (A+W) = 780 
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substituting for TW and TE makes the equations become:
2 * (A-W) = 780 (original first equation)
1.5 * (A+W) = 780 (original second equation)
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since both equations equal 780, then they are equal to each other (transitive property of algebraic operations).
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2 * (A-W) = 1.5 * (A+W)
removing parentheses:
2*A - 2*W = 1.5*A + 1.5*W
subtracting 1.5*A from both sides of equation and adding 2*W to both sides of equation:
2*A - 1.5*A = 1.5*W + 2*W
combining like terms:
.5*A = 3.5*W
multiplying both sides of equation by 2:
A = 7*W
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original first equation is:
2 * (A-W) = 780
removing parentheses:
2*A - 2*W = 780
substituting 7*W for A:
2*7*W - 2*W = 780
simplifying and combining like terms:
12*W = 780
dividing both sides by 12:
W = 65
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original second equation is:
1.5 * (A+W) = 780
removing parentheses:
1.5*A + 1.5*W = 780
substituting 65 for W:
1.5*A + 1.5*65 = 780
simplifying:
1.5*A + 97.5 = 780
subtracting 97.5 from both sides of equation:
1.5*A = 780 - 97.5
simplifying:
1.5*A = 682.5
dividing both sides of equation by 1.5
A = 682.5/1.5
simplifying:
A = 455
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original first and second equations are:
2 * (A-W) = 780 (original first equation)
1.5 * (A+W) = 780 (original second equation)
substituting 455 for A and 65 for W in original first equation:
2 * (455-65) = 780
simplifying:
2*390 = 780
780 = 780
first equation is good
substituting 455 for A and 65 for W in original second equation:
1.5 * (455+65) = 780
simplifying:
1.5 * (520) = 780
780 = 780
second equation is good.
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answer is:
Airplane speed is 455 miles per hour.
Wind speed is 65 miles per hour.