Let's define the three consecutive odd integers as:

• First odd integer: $x$
• Second odd integer: $x + 2$
• Third odd integer: $x + 4$

The problem states:

• Twice the first integer: $2x$
• Three times the second integer: $3(x + 2)$
• Twice the third integer: $2(x + 4)$

The equation provided is that the sum of twice the first and three times the second is 55 more than twice the third. Therefore, we can write:

$2x + 3(x + 2) = 2(x + 4) + 55$

Now, let's simplify and solve the equation:

$2x + 3x + 6 = 2x + 8 + 55$

Combine like terms:

$5x + 6 = 2x + 63$

Subtract $2x$ from both sides:

$3x + 6 = 63$

Subtract 6 from both sides:

$3x = 57$

Divide by 3:

$x = 19$

So the three consecutive odd integers are $x = 19$, $x + 2 = 21$, and $x + 4 = 23$.

Verification:

• Twice the first: $2 \times 19 = 38$
• Three times the second: $3 \times 21 = 63$
• Twice the third: $2 \times 23 = 46$

Check the sum:

$38 + 63 = 101$

Check if this equals 55 more than twice the third:

$46 + 55 = 101$

Since both sides are equal, the solution is correct.

The three consecutive odd integers are 19, 21, and 23.

Would you like any further details or have any questions? Here are 5 related questions:

1. How would the problem change if it involved even integers instead of odd ones?
2. Can you solve a similar problem where the integers are not consecutive?
3. What if the sum involved three times the first and twice the second instead?
4. How can this method be adapted to find four consecutive odd integers?
5. What happens if the sum condition is changed to a difference instead?

Tip: When solving word problems, defining variables clearly and step-by-step translation of the problem into an equation is key to avoiding confusion.