In each of the following questions, two equations are given. You have to solve them and state the correct relationship- $I. (3x^2 - 23x + 40 = 0)$ $II. (2y^2 - 23y + 66 = 0)$

Solving 3x² - 23x + 40 = 0 gives x = 5, 8/3 (approximately 5, 2.67). Solving 2y² - 23y + 66 = 0 gives y = 6, 11/2 (approximately 6, 5.5). Comparing: 5 < 6, 5 < 5.5, but 8/3 (2.67) < 5.5 and 2.67 < 6. Since some x values are NOT less than some y values (5 > 5.5 is false, but all x < all y is true). Actually checking: 5 < 5.5 ✓, 2.67 < 5.5 ✓, 5 < 6 ✓, 2.67 < 6 ✓. All x values are indeed less than all y values, so x < y is correct.