I. 4x2 + 9x - 9 = 0
II. 4y2 - 19y + 12 = 0
I. 4x2 + 9x - 9 = 0 => 4x2 + 12x – 3x – 9 = 0 => 4x(x + 3) – 3(x + 3) = 0 => (x + 3) (4x -3) = 0 => x = -3, 3/4 II. 4y2 - 19y + 12 = 0 => 4y2 - 16y – 3y + 12 = 0 => 4y(y - 4) – 3(y - 4) = 0 => (y - 4) (4y – 3) = 0 => y = 4, 3/4 Hence, x ≤ y.
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