Answer
$153 \text{mph}$.
Work Step by Step
The distance between the manufacturing plant and company headquarters is $300\,\text{mi}$.
The corporate pilot must fly from headquarters to the plant and back in $4\,\text{hr}$.
The speed of the air is $20-\text{mph}$ headwind and $20-\text{mph}$ tailwind.
Distance: $300$
Speed: $r-20$
Then put these values in the equation $\text{time}=\frac{\text{distance}}{\text{speed}}$.
Therefore,
$\begin{align}
& \text{time}=\frac{\text{distance}}{\text{speed}} \\
& \text{time}=\frac{\text{300}}{r\text{-20}}
\end{align}$
Distance: $300$
Speed: $r+20$
Then put these values in the equation $\text{time}=\frac{\text{distance}}{\text{speed}}$.
Therefore,
$\begin{align}
& \text{time}=\frac{\text{distance}}{\text{speed}} \\
& \text{time}=\frac{300}{r+20}
\end{align}$
Now form the equation:
$\begin{align}
& \frac{\text{300}}{r\text{-20}}+\frac{\text{300}}{r\text{+20}}=4 \\
& \frac{300\left( r-20 \right)+300\left( r+20 \right)}{\left( r-20 \right)\left( r\text{+20} \right)}=4 \\
& \frac{300r-6000+300r+6000}{\left( r-20 \right)\left( r\text{+20} \right)}=4 \\
& \frac{300r+300r}{\left( r-20 \right)\left( r\text{+20} \right)}=4
\end{align}$
Explain further:
$\begin{align}
\frac{300r+300r}{\left( r-20 \right)\left( r\text{+20} \right)}=4 & \\
\frac{1200r}{\left( {{r}^{2}}-{{20}^{2}} \right)}=4 & \\
\end{align}$
Multiply by $\left( {{r}^{2}}-{{20}^{2}} \right)$ on both sides:
$\begin{align}
& \frac{1200r}{\left( {{r}^{2}}-{{20}^{2}} \right)}\cdot \left( {{r}^{2}}-{{20}^{2}} \right)=4\cdot \left( {{r}^{2}}-{{20}^{2}} \right) \\
& 120r=4\cdot \left( {{r}^{2}}-{{20}^{2}} \right) \\
& r\approx 153 \text{mph}
\end{align}$
Therefore the plane has a speed of $r\approx 153 \text{mph}$
