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4 December, 01:48

Find four consecutive odd integers such that 3 times the largest decreased by 19 is twice the second integer

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  1. 4 December, 01:57
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    n, n + 2, n + 4, n + 6 - four consecutive odd integers

    3 times the largest decreased by 19 is twice the second integer

    3 (n + 6) - 19 = 2 (n + 2) use distributive property

    (3) (n) + (3) (6) - 19 = (2) (n) + (2) (2)

    3n + 18 - 19 = 2n + 4

    3n - 1 = 2n + 4 add 1 to both sides

    3n = 2n + 5 subtract 2n from both sides

    n = 5

    n + 2 = 5 + 2 = 7

    n + 4 = 5 + 4 = 9

    n + 6 = 5 + 6 = 11

    Answer: 5, 7, 9, 11
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