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

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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
• 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

• 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

• 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

• 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.

• 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?
• 135/3 = 45

43 + 45 + 47 = 135

• x + (x+2) + (x+4) = 135

3x + 6 = 135

x = 43

Verify : 43 + 45 + 47 = 135

• 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

• 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