I. x2 - 11x + 24 = 0
II. y² - 5y + 6 = 0
I. x2– 11x + 24 =0 => x2- 8x - 3x + 24 = 0 => x(x - 8) - 3(x - 8) = 0 => (x - 8) (x - 3) = 0 => x = 8, 3 II. y² - 5y + 6 = 0 => y2- 2y - 3y + 6 = 0 => y(y - 2) - 3(y - 2) = 0 => (y - 2) (y - 3) = 0 => y = 2, 3 Hence, x ≥ y.
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