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21 January, 12:25

find three consecutive even integers so that the twice the sum of the second and third is twelve less than six times the first

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  1. 21 January, 12:36
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    n, n + 2, n + 4 - three consecutive even integers

    the twice the sum of the second and third: 2[ (n + 2) + (n + 4) ]

    twelve less than six times the first: 6n - 12

    The equation:

    2[ (n + 2) + (n + 4) ] = 6n - 12

    2 (n + 2 + n + 4) = 6n - 12

    2 (2n + 6) = 6n - 12 use distributive property

    (2) (2n) + (2) (6) = 6n - 12

    4n + 12 = 6n - 12 subtract 12 from both sides

    4n = 6n - 24 subtract 6n from both sides

    -2n = - 24 divide both sides by (-2)

    n = 12

    n + 2 = 12 + 2 = 14

    n + 4 = 12 + 4 = 16

    Answer: 12, 14, 16
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