What's the solution of find the three consecutive odd integers whose sum is 135?

14 Answers

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  • 1 decade ago
    Best Answer

    First do 135 divided by 3, which is 45 then you take two off the 45 which is 43. Also add 2 to the 45 to make 47. So 43+45+47= 135

    Source(s): Math Geek
  • pfifer
    Lv 4
    3 years ago

    i'm going to help you with the 1st one. which will, in turn, help you with the subsequent 3. So: permit x be the 1st integer. Then x+a million is the subsequent consecutive integer, outstanding? And their sum is x + (x+a million). so which you're able to sparkling up the equation: x + (x+a million) = 7^2 2x + a million = 40 9 2x = 40 8 x = 24 so x+a million = 25; to that end, 24 and 25 is the answer to the 1st subject. For #5, you're able to translate this word subject into algebra. back, x is the 1st integer, x+a million the 2nd. Ten below two times the smaller is 2x - 10. So the equation you desire is: x+a million = 2x-10 11 = x and 12 = x+a million

  • 1 decade ago

    I would divide 135 by 3 to get an average and it came out to 45 then since you are looking for consecutive odd integers I would take the one below and the one above the average which would make it 43,45,47

  • maussy
    Lv 7
    1 decade ago

    The first is odd so we can write it as x+1

    if x+1 is the smallest than the others are x+3 and x+5

    Their sum is x+1+x+3+x+5 = 3x+9

    so x=42

    the integers are 43 , 45 , 47

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  • Max
    Lv 6
    1 decade ago

    x=the first odd integer

    x+2=the second odd integer

    x+4=the third odd integer

    x+x+2+x+4=135

    3x+6=135

    3x+6-6=135-6

    3x/3=129/3.x=43-the first integer. x+2=43+2=45 - the second integer..x+4=43+4=47-the third integer.

    43+45+47=135.

  • 1 decade ago

    let the 1st integer be = x

    let the 2nd integer be = x + 2

    let the 3rd integer be = x + 4

    now,solve:

    x + x + 2 + x + 4 = 135

    3x + 6 = 135

    transpose, 6 will be negative:

    3x = 135 - 6

    3x = 129

    divide both sides by 3:

    x = 43

    substitute:

    1st integer : x = 43

    2nd integer: 43 + 2 = 45

    3rd integer: 43 + 4 = 47

    Source(s): HOW'S THAT?
  • Helmut
    Lv 7
    1 decade ago

    135/3 = 45

    43 + 45 + 47 = 135

  • 1 decade ago

    x + (x+2) + (x+4) = 135

    3x + 6 = 135

    x = 43

    Verify : 43 + 45 + 47 = 135

  • 1 decade ago

    X=1st odd integer

    X+2=2nd odd integer

    X+4=3rd odd integer

    135=sum of the three

    equation:

    X+X+2+X+4=135

    3X+6=135

    3X=135-6

    3X=129

    Divide both sides by 3

    X=43

    X+1=45

    X+2=47

  • 1 decade ago

    Using Model Method

    [XXXXXXX] - 2

    [XXXXXXX] {This is the middle Bar}

    [XXXXXXX] + 2

    Add up the above 3 Bars

    3 x [XXXXXXX] = 135

    Therefore the middle Bar is,

    [XXXXXXX] = 135/3

    [XXXXXXX] = 45

    Therefore the three consecutive odd integers are:

    43, 45 & 47

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