Question 239236
Against a head wind, Jeff computes his flight time for a trip of 2900 miles at 5 hours.
 The flight would take 4 hours and 50 minutes if the head wind were half as much.
 Find the head wind and the plane's air speed.
:
Let x = plane's air speed
Let y = headwind speed
and 
.5y = half the headwind speed
:
Write two dist equations; dist = time * speed
:
Equation 1
5(x - y) = 2900
Simplify, divide both sides by 5
x - y = 580
and
equation 2
4{{{5/6}}}(x - .5y) = 2900
which is
{{{29/6}}}(x - .5y) = 2900
Divide both sides by 29/6: results
x - .5y = 600
:
use elimination
x - .5y = 600
x - y = 580
--------------Subtraction eliminate x, find y
+.5y = 20
y = 40 mph is the head wind
then
x - 40 = 580
x = 580 + 40
x = 620 mph is the plane speed
:
:
Check solution in the 2nd original equation
{{{29/6}}}(620 - .5(40)) = 
{{{29/6}}}(620 - 20) = 
{{{29/6}}}(600) = 2900
29(100) = 2900