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Let the first integer be x, the second be x + 1, and the third be x + 2. So we have:

2x + 4(x + 1) = 2(x + 2) + 20

2x + 4x + 4 = 2x + 4 + 20

4x = 20

x = 5

x + 1 = 6

x + 2 = 7

Thus these numbers are 5, 6, and 7

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Q: Find three consecutive integers such that the sum of twice the first and 4 times that second is equal to 20 more that twice the third?
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