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

first number: x

second number x+2
third number : x+4

solve 2(x + x+2 ) = 3(x+4) - 1

To solve this problem, let's represent the consecutive odd integers as variables. Let's assume the first odd integer is "x".

According to the problem, the consecutive odd integers are x, x + 2, and x + 4, since they are consecutive and odd numbers are always two units apart.

Now, we can set up an equation based on the given information. Twice the sum of the first and second is equal to one less than three times the third:

2(x + (x + 2)) = 3(x + 4) - 1.

Let's solve this equation step by step:

2(2x + 2) = 3x + 12 - 1.

First, distribute 2 to both terms inside the parentheses:

4x + 4 = 3x + 11.

Next, combine like terms by subtracting 3x from both sides:

4x - 3x + 4 = 11.

This simplifies to:

x + 4 = 11.

Finally, solve for x by subtracting 4 from both sides:

x = 11 - 4.

Thus, x = 7.

Now that we know the value of x, we can find the three consecutive odd integers:

The first integer is x = 7.
The second integer is x + 2 = 7 + 2 = 9.
The third integer is x + 4 = 7 + 4 = 11.

So, the three consecutive odd integers that satisfy the given condition are 7, 9, and 11.