Solve algebraically using one variable: Find three consecutive odd integers such that the

product of the first integer and the third integer is equal to nine more than twelve times the middle integer.

Let the first integer be x-2, then middle integer is x and the third is x+2.

(x-2)(x+2) = 12x + 9

X^2 - 4 = 12x + 9

x^2 - 12x - 13 = 0

(x+1)(x-13) = 0

11 * 15 = 12*13 + 9

To solve this algebraically, let's represent the three consecutive odd integers as x, x+2, and x+4.

According to the problem, the product of the first and third integers is equal to nine more than twelve times the middle integer. We can write this as an equation:

x * (x + 4) = 12(x + 2) + 9

Now, let's solve this equation step by step:

1. Distribute 12 to x + 2:
x * (x + 4) = 12x + 24 + 9

2. Simplify the right side:
x^2 + 4x = 12x + 33

3. Move all terms to one side of the equation:
x^2 + 4x - 12x - 33 = 0

4. Combine like terms:
x^2 - 8x - 33 = 0

Now, we have a quadratic equation. To solve it, we can use factoring, completing the square, or the quadratic formula.

Let's solve it using the quadratic formula:

5. Use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -8, and c = -33.

x = (-(-8) ± √((-8)^2 - 4(1)(-33))) / (2(1))

6. Simplify inside the square root:
x = (8 ± √(64 + 132)) / 2

x = (8 ± √196) / 2

7. Simplify inside the square root further:
x = (8 ± 14) / 2

8. Simplify:
x = (8 + 14) / 2 or x = (8 - 14) / 2

x = 22 / 2 or x = -6 / 2

x = 11 or x = -3

So, the possible values for x are 11 and -3.

Let's substitute these values back into our original representation of the integers:

For x = 11: The three consecutive odd integers are 11, 13, and 15.

For x = -3: The three consecutive odd integers are -3, -1, and 1.

Therefore, we have found the two sets of three consecutive odd integers that satisfy the given condition.