SOLUTION: Darius went on a bike ride of 48 miles. He realized that if he had gone 6 mph faster, he would have arrived 18 hours sooner. How fast did he actually ride?

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Question 1030715: Darius went on a bike ride of 48 miles. He realized that if he had gone 6 mph faster, he would have arrived 18 hours sooner. How fast did he actually ride?
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
                RATE         TIME        DISTANCE

Actual          r            t           48

If            r+6          t-18          48


RT=D, basic travel rates rule. Variable assignments indicated in the table. form the system of equations and solve.

Say if you still need further help.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Darius went on a bike ride of 48 miles. He realized that if he had gone 6 mph faster, he would have arrived 18 hours sooner. How fast did he actually ride?
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slower speed DATA:
dist = 48 miles ; rate = x mph ; time = d/r = 48/x hrs
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faster speed DATA:
dist = 48 miles ;; rate = x+6 mph ; time = d/r = 48/(x+6) hrs
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slower time - faster time = 18 hr
48/x - 48/(x+6) = 18
48*(x+6) - 48x = 18x(x+6)
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48*6 = 18x^2 + 108x
18x^2 + 108x - 288 = 0
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9x^2 + 54x - 144 = 0
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(x-2)(9x +72) = 0
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Ans: x = 2 mph (slower speed)
x+6 = 8 mph (faster speed)
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Cheers,
Stan H.