y\(^2\) - 3y > 18 = y\(^2\) - 3y - 18 > 0 y\(^2\) - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0 = (y + 3) (y - 6) > 0 Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0 = y > -3, y > 6 By simply substituting the value of y into the inequality, y > -3 is invalid, and y < 6 is also invalid. Therefore, y < - 3 or y > 6