SOLUTION: Jake drives a tractor from one town to another, a distance of 130 kilometers. He drives 9 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast does

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Question 902629: Jake drives a tractor from one town to another, a distance of 130 kilometers. He drives 9 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast does he drive each way?
Found 2 solutions by Jc0110, MathTherapy:
Answer by Jc0110(165) About Me  (Show Source):
You can put this solution on YOUR website!
Let speed = v while time = t.
Speed=Distance%2FTime
Depart trip: v=130%2Ft --> Equation 1
t=130%2Fv --> Equation 2 ... Substitute into Equation 3.
Return trip: v%2B9=130%2F%28t-1%29 --> Equation 3
v%2B9=130%2F%28%28130%2Fv%29-1%29
v%2B9=130%2F%28%28130-v%29%2Fv%29
v%2B9=130%2A%28v%2F%28130-v%29%29
v%2B9=130v%2F%28130-v%29
%28v%2B9%29%28130-v%29=130v
130v-v%5E2%2B1170-9v=130v
v%5E2%2B130v-130v%2B9v-1170=0
v%5E2%2B9v-1170=0
%28v%2B39%29%28v-30%29=0
v=-39 or v=30 ... Substitute both v=-39 and v=30 into Equation 2.
t=130%2F-39 or t=130%2F30
t=-10%2F3 hours or t=13%2F3 hours
t=13%2F3 hours is acceptable, while t=-10%2F3 hours is unacceptable because time cannot be in negative.
Substitute t=13%2F3 into Equation 1 and 3 for speed of depart trip and speed of return trip respectively.
------------------------------------------------------------------------
Depart trip: v=130%2F%2813%2F3%29
Speed of depart trip, v=30km per hour
Therefore, speed of depart trip is 30 km per hour.
Return trip: v%2B9=130%2F%28%2813%2F3%29-1%29
Speed of return trip, v+9=130%2F%28%2813%2F3%29-1%29
Speed of return trip, v+9=39km per hour
Therefore, speed of return trip is 39 km per hour.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Jake drives a tractor from one town to another, a distance of 130 kilometers. He drives 9 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast does he drive each way?


Let average speed during outbound trip be S

Then average speed during inbound trip is: S + 9

Time equation derived from this: 130%2FS+=+130%2F%28S+%2B+9%29+%2B+1

130(S + 9) = 130(S) + S(S + 9) -------- Multiplying by LCD, S(S + 9)

130S+%2B+1170+=+130S+%2B+S%5E2+%2B+9S

130S+%2B+1170+=+S%5E2+%2B+139S

S%5E2+%2B+139S+-+130S+-+1170+=+0

S%5E2+%2B+9S+-+1170+=+0

(S - 30)(S + 39) = 0

S, or average speed during outbound trip = highlight_green%2830%29 km//h

Average speed during inbound trip: 30 + 9, or highlight_green%2839%29 km/h

You can do the check!! 



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