Question 197289
 Find three consecutive integers such that twice the smallest increased by the largest is 128.
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1st: x-1
2nd: x
3rd: x+1
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Equation:
2(x-1)+ (x+1) = 128
2x-2 + x + 1 = 128
3x = 129
x = 43
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1st = x-1 = 42
2nd = x = 43
3rd = x+1 = 44
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Cheers,
Stan H.