In the questions given below you have to find the value of x and y and then compare the value of x and y and give the answer related to this. $I. (3x^2 - 5x - 12 = 0), II. (3y^2 - 8y - 16 = 0)$

For equation I: 3x² - 5x - 12 = 0 factors to (3x+4)(x-3) = 0, giving x = -4/3 or x = 3. For equation II: 3y² - 8y - 16 = 0 factors to (3y+4)(y-4) = 0, giving y = -4/3 or y = 4. Comparing values: x and y both equal -4/3 in one case (x = y), but x = 3 < y = 4 in the other. Since x can be equal to y or less than y depending on which values we choose, the relation cannot be consistently established.