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30 November, 09:26

How do i solve the sum of three consecutive integers is 48 less than twice the smallest of the three integers?

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  1. 30 November, 12:34
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    So let's say the three consecutive integers are x-1, x, x+1.

    2 (x-1) = x-1+x+x+1+48

    2x-2=3x+48

    x=-50.

    The three integers will be - 49,-50,-51
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