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23 March, 11:46

Find three consecutive odd integers such that the largest decreased by three times the second is 47 less than the smallest

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  1. 23 March, 12:13
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    X = smallest y = second z = biggest

    x = x

    y = x + 2

    z = x + 4

    (x+4) - 3 (x+2) = x - 47

    x+4 - 3x - 6 = x - 47

    -2x - 2 = x - 47

    -3x - 2 = - 47

    -3x = - 45

    x = 15

    x = 15

    y = 15 + 2

    z = 15 + 4

    x = 15

    y = 17

    z = 19
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