Problem 2:
We can solve this problem by analyzing the factored expressions and using casework to explore the possible values of x and y.
Factoring the Expressions:
(2x^2 + 4x + 8) can be factored as 2(x^2 + 2x + 4). We can further factor (x^2 + 2x + 4) by noticing it's a perfect square trinomial. So, 2(x^2 + 2x + 4) = 2(x + 2)^2.
(3y^2 - 6y + 7) can be factored as 3(y^2 - 2y + 7/3). This trinomial cannot be factored further with integer coefficients.
Setting Up the Cases: The equation is: 2(x + 2)^2 * 3(y^2 - 2y + 7/3) = 24
We can rewrite it as: (x + 2)^2 * (y^2 - 2y + 7/3) = 4 ---------(1)
Since 4 can be factored as 1 * 4 or 2 * 2, we will consider two cases based on these factorizations:
Case 1: (x + 2)^2 = 1 and (y^2 - 2y + 7/3) = 4
Case 2: (x + 2)^2 = 4 and (y^2 - 2y + 7/3) = 1
Solving Case 1:
(x + 2)^2 = 1 --> x + 2 = ±1 --> x = -1 or x = -3
(y^2 - 2y + 7/3) = 4 --> This quadratic factors as (y - 1)(y - 7/3) = 0. So, y = 1 or y = 7/3.
Therefore, in Case 1, we have four possible ordered pairs: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).
Solving Case 2:
(x + 2)^2 = 4 --> x + 2 = ±2 --> x = 0 or x = -4
(y^2 - 2y + 7/3) = 1 --> This quadratic has no real number solutions because the discriminant (b^2 - 4ac) is negative.
Therefore, Case 2 has no real number solutions for (x, y).
Total Ordered Pairs: Combining the results from both cases, we have a total of 4 possible ordered pairs from Case 1: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).
Answer: There are 4 possible ordered pairs (x, y) that satisfy the given equation.