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22 September, 15:27

find 3 consecutive integers such that twice the greatest integer is 2 less than 3 times the least integer

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  1. 22 September, 16:52
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    The consecutive integers are n, n+1, n+2

    2 (n+2) = 3n-2 perform indicated multiplication on left side

    2n+4=3n-2 subtract 3n from both sides

    -n+4=-2 subtract 4 from both sides

    -n=-6 divide both sides by - 1

    n=6

    So the three consecutive numbers are 6,7, and 8
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