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Answer :
The speed of the slower train is approximately 71 mi/h, and the speed of the faster train is 85 mi/h.
How did we get the value?
Let's assume the speed of the slower train is x mi/h.
Since the faster train is traveling 14 mi/h faster, its speed would be (x + 14) mi/h.
The combined speed of the two trains is the sum of their individual speeds. So, the equation can be formed as follows:
3(x + x + 14) = 468
Simplifying the equation, we have:
3(2x + 14) = 468
6x + 42 = 468
6x = 468 - 42
6x = 426
x = 426/6
x ≈ 71
Therefore, the speed of the slower train is approximately 71 mi/h, and the speed of the faster train is 71 + 14 = 85 mi/h.
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The complete question goes thus:
Two trains leave towns 468 mi apart at the same time and travel toward each other. One train travels 14 mi/h faster than the other. If they meet in 3 hours, what is the rate of each train?
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