SOLUTION: Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.

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Question 162521: Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
Answer by nerdybill(7384) About Me  (Show Source):
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Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
.
Let x = first (smallest) consecutive integer
x+1 = second consecutive integer
x+2 = third consecutive integer
x+3 = fourth (largest) consecutive integer
.
2[(x+2)+(x+3)] = 3x+91
2[2x+5] = 3x+91
4x+10 = 3x+91
x+10 = 91
x = 81
.
Therefore, the integers are:
81, 82, 83, and 84