Question 684837
A passenger plane made a trip to Las-Vegas and back. 

On the trip there it flew {{{s=400 mph}}}

 and on the return trip it went {{{s[1]=480 mph}}}. 

 the trip there: {{{d=s*(t+1h)}}}...it took one hour more for the plane to fly

the return trip: {{{d=s[1]*t}}}

we know that distance {{{d}}} is same in both directions; so, we have

{{{s*(t+1h)=s[1]*t}}}....plug in  {{{s=400 mph}}} and {{{s[1]=480 mph}}}.

{{{400 mph*(t+1h)=480 mph*t}}}.....solve for {{{t}}}

{{{400 mph*t+400 mph*1h=480 mph*t}}}

{{{400 mph*t-480 mph*t+400 mph^2=0}}}

{{{(400 mph-480 mph)t+400 mph^2=0}}}

{{{-80 mph*t=-400 mph^2}}}

{{{t=(-400 mph^2)/(-80 mph)}}}

{{{t=5h}}}...............now we can find the distance


{{{d=s[1]*t}}}

{{{d=480 (mil/h)*5h}}}

{{{d=480 (mil/cross(h))*5cross(h)}}}

{{{d=2400mil}}}

check the other direction:

{{{d=s*(t+1h)}}}

{{{d=400 (mil/h)*6h}}}

{{{d=400 (mil/cross(h))*6cross(h)}}}

{{{d=2400mil}}}