Three consecutive integers are such that the sum of the first and twice the second is 15 more than twice the third. Form a equation & solve?

This is for Algebra 1 honors (8th grade) and I need help asap. I am supposed to show all work but I have no idea how to.

6 Answers

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  • Como
    Lv 7
    3 weeks ago

    :-

    x + 2 ( x + 1 ) = 2 ( x + 2 ) + 15

    3x + 2 = 2x + 19

    x = 17

    Integers are 17 , 18 , 19

  • 3 weeks ago

    Three consecutive integers are such that

    the sum of the first and twice the second is 15 more than twice the third.

    Form a equation & solve?

    (x - 1) + 2x = 2(x + 1) + 15

    3x - 1 = 2x + 17

    x = 18

    The three consecutive integers are 17, 18, and 19.

  • TomV
    Lv 7
    3 weeks ago

    3 consecutive integers = n, n+1, and n+2

    first + twice the second = 15 more than twice the third

    n + 2(n+1) = 2(n+2)+15

    3n + 2 = 2n + 19

    3n - 2n = 19 - 2

    n = 17

    Ans:

     3 integers are 17, 18, and 19

  • david
    Lv 7
    3 weeks ago

    n = 1st,  n+1 2nd,  n+2 = 3rd

      n + 2(n+1) = 15 + 2(n+2)

      n + 2n + 2 = 15 + 2n + 4

    n + 2n - 2n = 15 + 4 - 2

       n = 17

      n +1 = 18

      n + 2 = 19

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  • alex
    Lv 7
    3 weeks ago

    Hint:

    Three consecutive integers are x , x+1 , x+2

    Form an equation then solve

  • x - 1 , x , x + 1

    Those are your numbers

    (x - 1) + 2 * x = 15 + 2 * (x + 1)

    Expand, combine like terms and solve for x

    x + 2x - 1 = 15 + 2x + 2

    3x - 1 = 17 + 2x

    3x - 2x = 17 + 1

    x = 18

    18 - 1 , 18 , 18 + 1

    17 , 18 , 19

    Those are your numbers.  Test it out

    17 + 2 * 18 = 17 + 36 = 53

    19 * 2 + 15 = 38 + 15 = 53

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