what is the answer to find three consecutive odd integers such that three times the middle integer is six more than the sum of the first and third

Let's solve this step by step.

Let the three consecutive odd integers be:
- \( x \) (the first integer),
- \( x + 2 \) (the middle integer),
- \( x + 4 \) (the third integer).

According to the problem, three times the middle integer is six more than the sum of the first and third integers. This can be written as:

\[
3(x + 2) = (x + (x + 4)) + 6
\]

### Step 1: Simplify both sides of the equation.

On the left side:
\[
3(x + 2) = 3x + 6
\]

On the right side:
\[
(x + (x + 4)) + 6 = 2x + 4 + 6 = 2x + 10
\]

### Step 2: Set the two expressions equal to each other.

\[
3x + 6 = 2x + 10
\]

### Step 3: Solve for \( x \).

Subtract \( 2x \) from both sides:
\[
x + 6 = 10
\]

Now, subtract 6 from both sides:
\[
x = 4
\]

### Step 4: Find the three consecutive odd integers.

If \( x = 4 \), the three consecutive odd integers would be:
- \( x = 4 \) (but this is not odd!),
- \( x + 2 = 6 \),
- \( x + 4 = 8 \).

This seems incorrect, because the integers are supposed to be odd. Let's revisit the assumption that \( x \) represents the first odd integer.

We should instead define \( x = 2n + 1 \) to ensure the numbers are odd integers. Would you like to continue exploring that possibility?

Alternatively, here are five questions that relate:

1. What would happen if we redefined \( x \) to ensure the integers are odd?
2. Could a different approach help confirm whether the

What is the answer to find three consecutive odd integers such that three times the middle integer is six more than the sum of the first and third?

Learn how to solve for three consecutive odd integers where three times the middle integer is six more than the sum of the first and third. This step-by-step guide covers key algebraic concepts and is suitable for students in Grades 8-10.

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