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.
is result