Find three consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third.

Question

Find three consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third.

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Thiên Hương 4 years 2021-08-31T12:13:20+00:00 2 Answers 9 views 0

Answers ( )

  1. Edana
    0
    2021-08-31T12:15:07+00:00
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    Answer:

    -19.333…

    -18.333…

    -17.333…

    Step-by-step explanation:

    First is x

    Second is x+1

    Third is x+2

    2x+3(x+1)=2(x+2)-55

    2x+3x+3=2x+4-55

    5x+3=2x-51

    5x-2x=-51-3

    3x=-58

    3x/3=-58/3

    x=-19.333…

    x+1=-19.333+1

    =-18.333…

    x+2=-19.333+2

    =-17.333…

  2. Farah
    0
    2021-08-31T12:15:16+00:00
    Reply

    Answer:

    Step-by-step explanation:

    let the numbers be n,n+2,n+4

    2n+3(n+2)=2(n+4)+55

    2n+3n+6=2n+8+55

    2n+3n-2n=63-6

    3n=57

    n=57/3=19

    so numbers are 19,21,23

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )