Question 131554: Find two consecutive odd integers such that seven times the first equals five times the second. Found 2 solutions by stanbon, solver91311:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find two consecutive odd integers such that seven times the first equals five times the second.
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Odd integers are always one more that an even integer:
1st: 2n+1
2nd: 2n+3
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EQUATION:
7(2n+1) = 5(2n+3)
14n+7 = 10n+15
4n = 8
n = 2
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1st: 2n+1 = 5
2nd: 2n+3 = 7
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Cheers,
Stan H.
You can put this solution on YOUR website! If an odd integer is x, then the next consecutive odd integer must be x + 2 (that's because x + 1 would be an even integer).
We are told that seven times the first [7x] is equal to five times the second [5(x + 2)].
So the first odd integer is 5 and the next consecutive one is 5 + 2 = 7, and by the way, 7 times 5 = 5 times 7, so the answer checks.
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