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)   (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)   (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.

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