Question 120519
Let r be the speed in still air, in accordance with rule one for word problems: Set the variable equal to the thing you want to know.


Air speed against the wind would then be {{{r-30}}} and airspeed with the wind would be {{{r+30}}}.  If the against the wind part of the trip takes t hours, then the return trip must take 10 - t hours, because the entire trip took 10 hours.


The formula for straight-line distance in terms of constant rate over a period of time is {{{d=rt}}}.  So we can express time as a function of distance and rate by saying {{{t=d/r}}}.


In our case, the time for the outbound trip, t, is given by {{{t=720/(r-30)}}}.  The time for the return trip, 10 - t, is given by {{{10-t=720/(r+30)}}}


First solve this second equation for t:


{{{-t=(720/(r+30))-10}}}


{{{t=10-(720/(r+30))}}}


{{{t=(10(r+30)-720)/(r+30)}}}


{{{t=(10r+300-720)/(r+30)}}}


{{{t=(10r-420)/(r+30)}}}


Now, note that we have two expressions for t in terms of r.  Set them equal to each other:


{{{720/(r-30)=(10r-420)/(r+30)}}}


Multiply by {{{(r+30)(r-30)}}}


{{{720(r+30)=(10r-420)(r-30)}}}


Distribute and collect terms


{{{720r+21600=10r^2-300r-420r+12600}}}
{{{10r^2-1440r-9000=0}}}


Divide by 10


{{{r^2-144r-900=0}}}


Note that {{{-150*6=-900}}} and {{{-150+6=-144}}}, so


{{{(r-150)(r+6)=0}}} => {{{r=150}}} or {{{r=-6}}}


The solution set to the quadratic is therefore 150 or -6.  Since flying backwards doesn't make much sense, exclude the -6 value as an extraneous root.  The speed in still air is 150 miles per hour.


Check:

{{{720/(150-30)=720/120=6}}} and {{{720/(150+30)=720/180=4}}}, and finally, 6 + 4 = 10.  Answer checks.


Hope that helps,
John