Answer
$22500 \text{ miles}$
Work Step by Step
Let $x$ be the time of the first rocket. Then $x-\dfrac{1}{4}$ is the time of the second rocket.
Using $D=rt,$ then the conditions of the problem for the first rocket is
\begin{array}{l}\require{cancel}
D_1=9000(x)
\\\\
D_1=9000x
.\end{array}
Using $D=rt,$ then the conditions of the problem for the second rocket is
\begin{array}{l}\require{cancel}
D_2=10000\left(x-\dfrac{1}{4}\right)
\\\\
D_2=10000x-2500
.\end{array}
Since the distances from earth are the same, then
\begin{array}{l}\require{cancel}
D_1=D_2
\\\\
9000x=10000x-2500
.\end{array}
Using the properties of equality, then
\begin{array}{l}\require{cancel}
9000x=10000x-2500
\\\\
9000x-10000x=-2500
\\\\
-1000x=-2500
\\\\
x=\dfrac{-2500}{-1000}
\\\\
x=2.5
.\end{array}
Hence, the distance from earth, $9000x,$ is $
22500 \text{ miles}
.$
