Answer:
11 mph and 20 mph
Step-by-step explanation:
Represent his average speed going by r1 and his average speed returning by r2.  We know that r1 = r2 + 9.
Recall that distance = rate times time, so time = distance / rate.
Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).
Time spend returning was (280 mi) / r2.
The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs
Note that there is only one variable here:  r2.  Find r2, and then from r2, find r1:
Dividing all 3 terms by 14 hrs yields:
   20            20
---------- + ----------- = 1
 r2 + 9         r2
The LCD here is r2(r2 + 9).  Thus, we have:
       20r2                    (r2 +  9)(r2)
------------------- = 1 or  ------------------
  (r2 +  9)(r2)               (r2 +  9)(r2)
Then 20(r2) = (r2)^2 + 9(r2).  This is reducible by dividing all terms by r2:
20 = r2 + 9, or 11 = r2.  Then r1 = 11 + 9, or 20.
The two rates were 11 mph and 20 mph.