Jeff recently drove to visit his parents who live 780 miles away. On his way there his average speed was 11 miles per hour faster than on his way home (he ran into some bad weather). If Jeff spent a total of 26 hours driving, find the two rates.

66 mph there, 55 mph back

Step-by-step explanation:

Each leg of the trip he drove 780 miles. The rate going there was faster that the rate coming home, so we have two equations based on distance ... This would also effect the times going there and back, so we have two times as well ...

Distance = (rate) (time) so

Going there ...

780 = (r + 11) (t1)

and coming back ...

780 = (r) (t2)

We know that t1 + t2 = 26, so solve both equations for t, then add them together ...

780 / (r + 11) = t1 and 780/r = t2

780 / (r + 11) + 780/r = t1 + t2

780 / (r + 11) + 780/r = 26

now solve for r

780r + 780 (r + 11) = 26[r (r + 11) ] (multiply both sides by r (r + 11) to get rid of the fractions)

Simplify ...

780r + 780r + 8580 = 26r² + 286r

We now have a quadratic, so get everything to one side, and solve ...

26r² - 1274r - 8580 = 0

r² - 49r - 330 = 0 (divide both sides by 26)

(r - 55) (r + 6) = 0 (factor)

So

r - 55 = 0, then r = 55

and

r + 6 = 0, then r = - 6 * **This answer is ignored because we are talking about rate, and rate in regards to speed is always positive, so

r = 55 is the only acceptable answer.

His rate going there was 66 mph, and his rate coming back was 55 mph