SOLUTION: Jessie recently drove to visit her parents who live 756 miles away. On her way there her average speed was 21 miles per hour faster than on her way home (she ran into some bad weat

Question 1056787: Jessie recently drove to visit her parents who live 756 miles away. On her way there her average speed was 21 miles per hour faster than on her way home (she ran into some bad weather). If Jessie spent a total of 21 hours driving, find the two rates.
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52841) (Show Source): You can put this solution on YOUR website!
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Jessie recently drove to visit her parents who live 756 miles away. On her way there her average speed was 21 miles per hour faster
than on her way home (she ran into some bad weather). If Jessie spent a total of 21 hours driving, find the two rates.
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Let x be the slower rate, in miles per hour (mph). Then the faster rate is (x+21) mph. Jessie spent hours for the way "there". She spent hours for the returning trip. The condition says = 21. Solve for x. For it, multiply both sides by x*(x+21). You will get 756x + 756*(x+21) = 21x*(x+1). Divide both sides by 21. You will get 36x + 36(x+21) = x*(x+1). Simplify and solve this quadratic equation.

Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
Jessie recently drove to visit her parents who live 756 miles away. On her way there her average speed was 21 miles per hour faster than on her way home (she ran into some bad weather). If Jessie spent a total of 21 hours driving, find the two rates.

Let speed to the house be S
Then return-trip speed = S - 21
We then get the following TIME equation:
756(S - 21) + 756S = 21S(S - 21) --- Multiplying by LCD, S(S - 21)
---------- Dividing both sides by 21

S(S - 84) - 9(S - 84) = 0
(S - 84)(S - 9) = 0
S, or speed to the house = OR S = 9 (ignore)
Speed on return trip: 84 - 21, or
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