SOLUTION: Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was h

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was h      Log On

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Question 1200264: Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was her speed from Tortula to Cactus?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Tortula to Cactus r, 204 miles
Cactus to Dry Junc. r+13, 308 miles
Total time, 6 hours

First part's time, 204%2Fr
Second part's time, 308%2F%28r%2B13%29

TIME SUM: 204%2Fr%2B308%2F%28r%2B13%29=6-----------Solve for r.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was her speed from Tortula to Cactus?

Let speed from Tortula to Cactus, be S
Then speed from Cactus to Dry Junction is S + 13
Since it took 6 hours to complete the entire journey, we get TIME equation: matrix%281%2C3%2C+6%2C+%22=%22%2C+204%2FS+%2B+308%2F%28S+%2B+13%29%29
                     matrix%281%2C3%2C+3%2C+%22=%22%2C+102%2FS+%2B+154%2F%28S+%2B+13%29%29 ---- Factoring out GCF, 2, in numerators
            3S(S + 13) = 102(S + 13) + 154S ----- Multiplying by LCD, S(S + 13) 

3S(S - 78) + 17(S - 78) = 0
S - 78 = 0             or             3S + 17 = 0____3S = - 17 (ignore)

Speed from Tortula to Cactus, or S = 78 mph