Distance/Rate/Time word problem

[acen_gr]

Homework Statement

A Cessna 150 averages 150 mph in still air. With a tailwind it is able to make a trip in 2 1/3 hours. Because of the headwind, it is only able to make the return trip in 3 1/2 hours. What is the average wind speed?

Homework Equations

(1) If x = wind speed and y = still air speed
∴ resultant speed (tailwind) = x + y
∴ resultant speed (headwind) = x - y

(2) d = rt

The Attempt at a Solution

let x = wind speed
still air speed = 150 mph
tailwind time = 2 1/3 hours
tailwind speed = 150 + x
headwind time = 3/12 hours
headwind speed = 150 - x

I used these as given for the problem, expressed distance in terms of tailwind/headwind time and rates, set distance as constant (equates), solves for x then gets 57.69 as answer.
but in the book, the answer is 30 mph. What is the problem with my solution?

Thanks in advance.

 

[HallsofIvy]

Well, your solution is wrong! And we can tell you what you did wrong because you do not say what you did. Please write out the equations exactly as you used them and how you solved.

 

[acen_gr]

let x = wind speed
still air speed = 150 mph
tailwind time = 2 1/3 hours
tailwind speed = 150 + x
headwind time = 3 1/2 hours
headwind speed = 150 - x

d = rt

tailwind:
d = (150 + x)(2 1/3)
d = 100 + 2 1/3 x

headwind:
d = (150 - x)(3 1/2)
d = 225 - 3 1/2 x

d = d

100 + 2x/3 = 225 - 3x/2
2x/3 + 3x/2 = 125
4x + 9x = 125(6)
13x = 750
x = 57.69

@HallsOfIvy, sorry..

 

[ehild]

let x = wind speed
still air speed = 150 mph
tailwind time = 2 1/3 hours
tailwind speed = 150 + x
headwind time = 3 1/2 hours
headwind speed = 150 - x

d = rt

tailwind:
d = (150 + x)(2 1/3)
d = 100 + 2 1/3 x

headwind:
d = (150 - x)(3 1/2)
d = 225 - 3 1/2 x

d = d

100 + 2x/3 = 225 - 3x/2
2x/3 + 3x/2 = 125
4x + 9x = 125(6)
13x = 750
x = 57.69

Note that 3 1/2=3.5=7/2 and 2 1/3 =7/3.

ehild