For three consecutive odd integers, the sum of 3 times the first integer and 2 times the second integer is 11 less than 4 times the third integer. What are the three integers?

Question

For three consecutive odd integers, the sum of 3 times the first integer and 2 times the second integer is 11 less than 4 times the third integer. What are the three integers?

Separate your answers with commas. Ex: 1,2,3

Answer 1

let the three consecutive odd integers be

x, x+2, and x+4

3x + 2(x+2) = 4(x+4) - 11

solve for x and sub into my definitions

Answer 2

To solve this problem, let's represent the three consecutive odd integers:

Let x be the first odd integer.
The next odd integer would be x + 2.
The third odd integer would be x + 4.

According to the given information, the sum of 3 times the first integer and 2 times the second integer is 11 less than 4 times the third integer:

3x + 2(x + 2) = 4(x + 4) - 11

Now, we can solve this equation step by step:

3x + 2(x + 2) = 4(x + 4) - 11
3x + 2x + 4 = 4x + 16 - 11
5x + 4 = 4x + 5
5x - 4x = 5 - 4
x = 1

So, the first odd integer is 1.
The second odd integer is x + 2 = 1 + 2 = 3.
The third odd integer is x + 4 = 1 + 4 = 5.

Therefore, the three consecutive odd integers are 1, 3, and 5.

For three consecutive odd​ integers, the sum of 3 times the first integer and 2 times the second integer is 11 less than 4 times the third integer. What are the three​ integers?

let the three consecutive odd integers be

To solve this problem, let's represent the three consecutive odd integers:

For three consecutive odd integers, the sum of 3 times the first integer and 2 times the second integer is 11 less than 4 times the third integer. What are the three integers?