In each of the following questions two equations (I) and (II) are given. Solve these equations and give the answer: ( (I) x^2 - 4x - 12 = 0 ) ( (II) 3y^2 - 8y + 5 = 0 )

Equation (I): x² - 4x - 12 = 0. Solving: x = [4 ± √(16 + 48)]/2 = [4 ± √64]/2 = [4 ± 8]/2. So x = 12/2 = 6 or x = -4/2 = -2. Equation (II): 3y² - 8y + 5 = 0. Solving: y = [8 ± √(64 - 60)]/6 = [8 ± √4]/6 = [8 ± 2]/6. So y = 10/6 = 5/3 or y = 6/6 = 1. Comparing: x = {-2, 6} and y = {1, 5/3}. We have: -2 < 1, -2 < 5/3, 6 > 1, 6 > 5/3. Since x is sometimes less than y and sometimes greater than y, no consistent relationship exists.