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14 September, 04:47

Find three consecutive odd integers such that 5 times the sum of all three is 42 more than the product of the first and second integers

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  1. 14 September, 06:06
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    The three integers are x, x+2, x+4

    5 (3x+6) = 42+x (x+2)

    15x+30=x^2+2x+42

    x^2 - 13x + 12=0

    (x-12) (x-1) = 0

    x = 12 or 1

    since the integers have to be odd we exclude 12

    integers are 1,3,5
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