SOLUTION: Justin recently drove to see his parents who live 390 miles away. on his way there his average speed was 11 mph faster than on his way home (he ran into some bad weather) if justin

Question 315203: Justin recently drove to see his parents who live 390 miles away. on his way there his average speed was 11 mph faster than on his way home (he ran into some bad weather) if justin spent a total of 13 hours driving find the two rates
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
Justin recently drove to see his parents who live 390 miles away. on his way there his average speed was 11 mph faster than on his way home (he ran into some bad weather) if justin spent a total of 13 hours driving find the two rates
Distance D = Rate R * Time T
d1 = distance to get there, r1 = rate to get there, t1 = time to get there
d2 = distance to get back, r2 = rate to get back, t2 = time to get back
d1 = d2 = 390 miles, r1 = r2 + 11 mph, t1 = 13 hours - t2
way there to get r2:
390 = (r2 + 11) * (13 - t2)
390/(13 - t2) = r2 + 11
390/(13 - t2) - 11 = r2
(390 - 11(13 - t2))/(13 - t2) = r2
(390 - 143 + 11t2)/(13 - t2) = r2
(11t2 + 247)/(13 - t2) = r2
way back to get t2:
390 = (11t2 + 247)/(13 - t2) * t2
390 = (11(t2)^2 + 247t2)/(13 - t2)
390(13 - t2) = 11(t2)^2 + 247t2
5070 - 390t2 = 11(t2)^2 + 247t2
-11(t2)^2 - 390t2 - 247t2 + 5070 = 0 (rearranged terms)
-11(t2)^2 - 637t2 + 5070 = 0
11(t2)^2 + 637t2 - 5070 = 0 (divided both sides by -1)

a = 11, b = 637, c = -5070

(the result that involves -637 - 793 is thrown out that would result in a negative answer)

t2 = 156/22 = 78/11 = 7 1/11 = 7.091 to 3 places hours
t1 = 13 - t2 = 143/11 - 78/11 = 65/11 = 5 10/11 = 5.909 to 3 places hours
390 = r1 * 65/11
(390 * 11)/65 = r1
4290/65 = 858/13 = 66 mph = r1
390 = r2 * 78/11
(390 * 11)/78 = r2
4290/78 = 2145/39 = 55 mph = r2
does r1 = r2 + 11? --> 55 + 11 = 66 --> yes

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