Jon recently drove to visit his parents who live 280 miles away. on his way there his average speed was 9 miles per hour faster than on his way home (he ran into some bad weather). if jon spent a total of 14 hours driving, find the two rates

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.

2x^2+5x+17

your welcome. : )