Answer
$135 \text{ mph}$
Work Step by Step
Let $x$ be the speed of the plane in still air.
Using $D=rt,$ then the conditions of the problem with the wind is
\begin{array}{l}\require{cancel}
465=(x+20)t
\\\\
\dfrac{465}{x+20}=t
.\end{array}
Using $D=rt,$ then the conditions of the problem against the wind is
\begin{array}{l}\require{cancel}
345=(x-20)t
\\\\
\dfrac{345}{x-20}=t
.\end{array}
Equating the two equations of $t$ and using the properties of equality result to
\begin{array}{l}\require{cancel}
\dfrac{465}{x+20}=\dfrac{345}{x-20}
\\\\
(x-20)(465)=(x+20)(345)
\\\\
465x-9300=345x+6900
\\\\
465x-345x=6900+9300
\\\\
120x=16200
\\\\
x=\dfrac{16200}{120}
\\\\
x=135
.\end{array}
Hence, the speed of the plane in still air, $x,$ is $
135 \text{ mph}
.$
