#### Answer

$6$ mph.

#### Work Step by Step

distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$
Let x be the rate of the home to campus leg.
The rate of the campus to home leg is $x-9.$
We are given:
total time = $\displaystyle \frac{5}{\text{rate of H to C}} + \displaystyle \frac{5}{\text{rate of C to H.}}=\frac{7}{6}$
That is,
$\displaystyle \frac{5}{x}+\frac{5}{x-9} =\frac{7}{6}\qquad$ ... $/\times 6x(x-9)$
$30(x-9)+30x=7x(x-9) \qquad$ ... distribute...
$ 30x-270+30x=7x^{2}-63x \qquad$ ... add $270-60x$
$0=7x^{2}-123x+270$
... Trying with synthetic division, we find:
$$\begin{array}{rrrrrr}
15 & | & 7 & -123 & 270 & \\
& & & 105 & -270 & \\
& & -- & -- & -- & \\
& & 7 & -18 & 0 &
\end{array}$$
$ 0=(x-15)(7x-18)\qquad$ ...
$x=15$ mph, or $\displaystyle \frac{18}{7}$ mph.
The rate of the return leg is $15-x:$
$15-9=6,$ or
$\displaystyle \frac{18}{7}-9$, which is negative , so we discard this possibility.
So, the rate of the return leg is $6$ mph.