Question 1208722
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A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes 5 hours and 20 minutes, 
but when there is a wind blowing from Penthaven to Jackson at 70 miles per hour, the trip takes 9 hours. 
How many miles is the distance from Penthaven to Jackson?
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<pre>
Let "d" be the distance from Penthaven to Jackson, in miles.

Let "u" be the speed of the plane in still air (at no wind), in miles per hour.


Since the round trip is 5 hours and 20 minutes, we can write an equation

    {{{(2d)/u}}} = 5{{{1/3}}}  hours,

or

    2d = {{{(16/3)u}}},   d = {{{(8u)/3}}}.


Next, the speed of the plane with the wind is u+70 mph  and the travel time with the wind is  

     {{{d/(u+70)}}} = {{{(((8u)/3))/(u+70)}}} = {{{(8u)/(3*(u+70))}}}.


Next, the speed of the plane against the wind is u-70 mph  and the travel time against the wind is  

     {{{d/(u-70)}}} = {{{(((8u)/3))/(u-70)}}} = {{{(8u)/(3*(u-70))}}}.


So, the time equation for the round trip with and against the wind is

    {{{(8u)/(3*(u+70))}}} + {{{(8u)/(3*(u-70))}}} = 9.


    +----------------------------------------------------------------------+
    |      At this point, the setup is complete.  Now our task is          |
    |  to solve this equation and to find the own speed "u" of the plane.  |
    +----------------------------------------------------------------------+


For it, multiply both sides of this equation by 3*(u+70*(u-70).  You will get

    8u(u+70) + 8u(u-70) = 9*3*(u+70)*(u-70),

    8u^2 + 560u + 8u^2 - 560u = 27*(u^2 - 4900),

    8u^2 + 8u^2 = 27u^2 - 27*4900,

   27*4900 = 27u^2 - 8u^2 - 8u^2

   27*4900 = 11u^2

    u = {{{sqrt((27*4900)/11)}}} = 109.669 miles per hour.


Thus the own speed of the plane at no wind is 109.669 miles per hour.


Hence, the distance from Penthaven to Jackson is d = {{{(8u)/3}}} = {{{(8*109.669)/3}}} = 292.451 miles.


<U>ANSWER</U>.   The distance from Penthaven to Jackson is about 292.5 miles.
</pre>

Solved.


The method, which I showed in this post, is a normal, regular and standard method solving similar problems.


On the way, I fixed the error which tutor/professor Edwin made in his post.


I do not think that a/the school teacher would accept the solution presented by Edwin.