An airplane takes \( 3 \) hours to travel a distance of \( 2700 \) miles with the wind. The return trip takes \( 5 \)hours against the wind. Find the speed of the plane in still air and the speed of the wind. The speed of the plane is \( \square \) mph and the speed of the wind is \( \square \) mph.
Real Tutor Solution
Quick Answer
\(x + y = \frac { 2700 } { 3 } , x - y = \frac { 2700 } { 5 } : x = 720 , y = 180 \)
Steps
\( \left [ \begin{array} { l } { x + y = \frac { 2700 } { 3 } } \\ { x - y = \frac { 2700 } { 5 } } \end{array} \right ] \)
Isolate \( x \) for \( x + y = \frac { 2700 } { 3 } : x = 900 - y \)
Substitute \( x = 900 - y \)
\( [ 900 - y - y = \frac { 2700 } { 5 } ] \)
Simplify
\( [ 900 - 2 y = 540 ] \)
Isolate \( y \) for \( 900 - 2 y = 540 : y = 180\)
For \( x = 900 - y \)
Substitute \( y = 180 \)
\( x = 900 - 180 \)
Simplify
\( x = 720 \)
The solutions to the system of equations are:
\( x = 720 , y = 180\)
The speed of the plane is 720 mph
The speed of the wind is 180 mph
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