Question 414660
You have 2 trips, each one is 756 mi.
Let {{{s}}} = the speed of the plane in still air
Let {{{w}}} = the wind speed
Speed with the wind = {{{s + w}}}
Speed against the wind = {{{s - w}}}
{{{d = 756}}} mi
----------------------
given:
With the wind:
(1) {{{756 = (s + w)*9}}}
against the wind:
(2) {{{756 = (s - w)*18}}}
-----------------
(1) {{{756 = (s + w)*9}}}
(1) {{{84 = s + w }}}
and
(2) {{{756 = (s - w)*18}}}
(2) {{{42 = s - w }}}
----------------
Add equations (1) and (2)
{{{126 = 2s}}}
{{{s = 63}}}
and
(2) {{{42 = s - w }}}
(2) {{{42 = 63 - w }}}
(2) {{{w = 63 - 42}}}
{{{w = 21}}}
The speed of the plane in still air is 63 mi/hr
The wind speed is 21 mi/hr
check answers:
(1) {{{756 = (s + w)*9}}}
(1) {{{756 = (63 + 21)*9}}}
(1) {{{756 = 84*9}}}
(1) {{{756 = 756}}}
and
(2) {{{756 = (s - w)*18}}}
(2) {{{756 = (63 - 21)*18}}}
(2) {{{756 = 42*18}}}
(2) {{{756  = 756}}}
OK
In practical terms, I don't think an airplane could
stay in the air doing only 63 mi/hr, but I think my method is good.