Answer
$15 \text{ } mph$
Work Step by Step
Let $x$ be the speed of the boat in still water. Using $D=rt,$ then, going downstream,
\begin{array}{l}\require{cancel}
20=(x+5)t
\\\\
\dfrac{20}{x+5}=t
.\end{array}
Using $D=rt,$ then, going upstream,
\begin{array}{l}\require{cancel}
10=(x-5)t
\\\\
\dfrac{10}{x-5}=t
.\end{array}
Equating the equations of $t,$ then
\begin{array}{l}\require{cancel}
\dfrac{20}{x+5}=\dfrac{10}{x-5}
.\end{array}
By cross-multiplication and using the properties of equality, then
\begin{array}{l}\require{cancel}
20(x-5)=10(x+5)
\\\\
20x-100=10x+50
\\\\
20x-10x=50+100
\\\\
10x=150
\\\\
x=\dfrac{150}{10}
\\\\
x=15
.\end{array}
Hence, the speed of the boat in still water, $x,$ is $
15 \text{ } mph
.$
