Let's call the speed on the way to campus "x". Since you ride 9 miles per hour faster on the way to campus than on the way home, the speed on the way home is "x - 9".
We know that the distance to campus and back is a total of 5 miles in each direction, so the total distance traveled is 10 miles.
We also know that the total time it takes is 7/6 hours.
Using the formula:
Total distance = (average speed) x (total time)
we can set up two equations:
5 = x(t1) (distance to campus)
5 = (x - 9)(t2) (distance back home)
where t1 is the time it takes to get to campus and t2 is the time it takes to get back home.
We also know that t1 + t2 = 7/6, since the total time is 7/6 hours.
We can solve for t1 in terms of t2:
t1 = (7/6 - t2)
Now we can substitute this value for t1 in the first equation:
5 = x(7/6 - t2)
Solving for x, we get:
x = 30/(7 - 6t2)
We want to find the speed on the way back home, which is x - 9.
x - 9 = 30/(7 - 6t2) - 9
Simplifying:
x - 9 = (30 - 9(7 - 6t2))/(7 - 6t2)
x - 9 = (3t2 - 3)/(-6t2 + 7)
Now we can use the fact that t1 + t2 = 7/6 to solve for t2:
t1 + t2 = 7/6
(7/6 - t2) + t2 = 7/6
Solving for t2:
t2 = 1/2
Now we can plug in t2 = 1/2 into the expression we found for x - 9:
x - 9 = (3(1/2) - 3)/(-6(1/2) + 7)
x - 9 = 3/5
x = 42/5
So the speed on the way back home is x - 9 = (42/5) - 9 = 3/5, which is approximately 0.6 miles per minute.