Answer
$12 \text{ miles}$
Work Step by Step
Let $x$ be the time of the first jogger. Then $x+0.5$ is the time of the second jogger.
Using $D=rt,$ then the conditions of the problem for the first jogger is
\begin{array}{l}\require{cancel}
D_1=8x
.\end{array}
Using $D=rt,$ then the conditions of the problem for the second jogger is
\begin{array}{l}\require{cancel}
D_2=6(x+0.5)
\\\\
D_2=6x+3
.\end{array}
Since the distances from the designated initial point are the same, then
\begin{array}{l}\require{cancel}
D_1=D_2
\\\\
8x=6x+3
.\end{array}
Using the properties of equality, then
\begin{array}{l}\require{cancel}
8x=6x+3
\\\\
8x-6x=3
\\\\
2x=3
\\\\
x=\dfrac{3}{2}
.\end{array}
Hence, the distance of the run, $8x,$ is $
12 \text{ miles}
.$
