Question 119180
What is common to both trips? That's always the right question to ask.
The answer is -distance-It's 255 mi in both cases. The speed will be
different and so will the time.
s = plane's average speed in still air
w = average windspeed
On the first trip, {{{255 = (s -w)1.7}}}
On the 2nd trip, {{{255 = (s + w)1.5}}}
{{{1.7s - 1.7w = 255}}}
{{{1.5s + 1.5w = 255}}}
Multiply both sides of the 2nd equation by the fraction {{{1.7/1.5}}}
{{{1.7s - 1.7w = 255}}}
{{{1.7s + 1.7w = 255*(1.7/1.5)}}}
add the equations
{{{3.4s = 255 + 255*(1.7/1.5)}}}
multiply both sides by 1.5
{{{5.1s = 255*1.5 + 255*1.7}}}
{{{5.1s = 382.5 + 433.5}}}
{{{5.1s = 876}}}
{{{s = 160}}}mi/hr average airspeed of plane
{{{1.7*160 - 1.7w = 255}}}
{{{272 - 1.7w = 255}}}
{{{1.7w = 272 - 255}}}
{{{1.7w = 17}}}
{{{w = 10}}}mi/hr ave windspeed
check answers
{{{1.7*160 - 1.7*10 = 255}}}
{{{272 - 17 = 255}}}
{{{255 = 255}}}
and
{{{1.5s + 1.5w = 255}}}
{{{1.5*160 + 1.5*10 = 255}}}
{{{240 + 15 = 255}}}
{{{255 = 255}}}
OK