<PRE><B>Question 833993</B>
Find 4 consecutive integers such that twice the sum of the first and third is 11 greater than 3 times the second. 
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1st: x-2
2nd: x-1
3rd: x
4th: x+1
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Equation:
2(x-2 + x) = 3(x-1)+11
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2(2x-2) = 3x-3+11
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4x-4 = 3x+8
x = 12
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1st: x-2 = 10
2nd: 11
3rd: 12
4th: 13
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Cheers,
Stan H.
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