SOLUTION: Kiran drove from Tortula to Cactus, a distance of 282 mi. She increased her speed by 9 mi/h for the 392 mi trip from Cactus to Dry Junction. If the total trip took 13 h, what was

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Kiran drove from Tortula to Cactus, a distance of 282 mi. She increased her speed by 9 mi/h for the 392 mi trip from Cactus to Dry Junction. If the total trip took 13 h, what was       Log On

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Question 988739: Kiran drove from Tortula to Cactus, a distance of 282 mi. She increased her speed by 9 mi/h for the 392 mi trip from Cactus to Dry Junction. If the total trip took 13 h, what was her speed from Tortula to Cactus?

I am coming up with an answer of something like 557 plus or minus 23 times square root of 337 all over 26.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Kiran drove from Tortula to Cactus, a distance of 282 mi. She increased her speed by 9 mi/h for the 392 mi trip from Cactus to Dry Junction. If the total trip took 13 h, what was her speed from Tortula to Cactus?
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T to C DATA:
distance = 282 mi : rate = r mph ; time = d/r = 282/r hours
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C to D DATA:
dist = 392 mi ; rate = r+9 mph ; time = 392/(r+9) hr
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Equation:
time + time = 13
282/r + 392(r+9) = 13
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282r + 282*9 + 392r = 13r(r+9)
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674r + 2538 = 13r^2 + 117r
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13r^2 - 557r - 2538 = 0
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Ans: r = 47 mph (rate from T to C)
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Cheers,
Stan H.
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