Answer
Speed of plane $=630$ mph.
Speed of wind $=90$ mph.
Work Step by Step
Step 1:- Assume unknown quantities as variables.
Let the rate of the plane in still air be $=x$.
Let the rate of the wind be $=y$.
Step 2:- Write the system of equations.
Speed of the plane with the wind $=(x+y)$.
Speed of the plane against the wind $=(x-y)$.
Formula of the distance: $speed \times time=distance $.
The given values are
Distance with the wind: $2160$ miles in $3$ hours.
Distance against the wind: $2160$ miles in $4$ hours.
Or we can write.
$\Rightarrow (x+y)\times 3=2160$ ...... (1)
$\Rightarrow (x-y)\times 4=2160$...... (2)
Step 3:- Solve the system of equations.
Divide equation (1) by $3$.
$\Rightarrow x+y=720$ ...... (3)
Divide the equation (2) by $4$.
$\Rightarrow x-y=540$...... (4)
Add equation (2) and (3).
$\Rightarrow x+y+x-y=720+540$
Simplify.
$\Rightarrow 2x=1260$
Divide both sides by $2$.
$\Rightarrow \frac{2x}{2}=\frac{1260}{2}$
Simplify.
$\Rightarrow x=630$
Plug the value of $x$ into equation (3).
$\Rightarrow 630+y=720$
Isolate $y$.
$\Rightarrow y=720-630$
Simplify.
$\Rightarrow y=90$.
Step 4:- Check the answers.
Substitute the values of $x$ and $y$ into equation (1).
$\Rightarrow (630+90)\times 3=2160$
$\Rightarrow 720\times 3=2160$
$\Rightarrow 2160=2160$ True.
