Solve the following equations and find the values of (x) and (y): I. (17x^2 + 48x = 9) II. (13y^2 = 32y - 12)

Solve equation I: 17x² + 48x - 9 = 0. Using quadratic formula, x = [-48 ± √(2304 + 612)]/34 = [-48 ± √2916]/34 = [-48 ± 54]/34. This gives x = 6/34 = 3/17 ≈ 0.176 or x = -102/34 = -3. Solve equation II: 13y² - 32y + 12 = 0. Using quadratic formula, y = [32 ± √(1024 - 624)]/26 = [32 ± √400]/26 = [32 ± 20]/26. This gives y = 52/26 = 2 or y = 12/26 = 6/13 ≈ 0.462. Comparing all values, both positive solutions (3/17 ≈ 0.176 and 6/13 ≈ 0.462) give x < y, and the negative solution x = -3 is also less than both positive y values. Therefore x < y always holds.