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8 April, 15:05

Find five consecutive integers such that the sum of the first and 5 times the third is equal to 41 less than 3 times the sum of the second fourth and fifth

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  1. 8 April, 15:24
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    see below

    Step-by-step explanation:

    We'll cal the first integer x and then the rest of them will be x + 1, x + 2, x + 3 and x + 4. We can write x + 5 (x + 2) = 3 (x + 1 + x + 3 + x + 4) - 41.

    x + 5x + 10 = 3 (3x + 8) - 41

    6x + 10 = 9x + 24 - 41

    6x + 10 = 9x - 17

    3x = 27

    x = 9

    The numbers are 9, 10, 11, 12, 13.
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