In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. $I. (6x^2 - 25x + 14 = 0) II. (9y^2 - 9y + 2 = 0)$

Solve equation I: 6x² - 25x + 14 = 0 factors to (2x-7)(3x-2) = 0, so x = 7/2 = 3.5 or x = 2/3 ≈ 0.67. Solve equation II: 9y² - 9y + 2 = 0 factors to (3y-1)(3y-2) = 0, so y = 1/3 ≈ 0.33 or y = 2/3 ≈ 0.67. Comparing values: When x = 3.5, it's greater than both y values (0.33, 0.67). When x = 0.67, it equals y = 0.67 and is greater than y = 0.33. Since x is always greater than or equal to y (with equality occurring when both are 2/3), the answer is x ≥ y.