I. x ² + 5 x + 6 = 0
II. y²+ 7 y + 12= 0
I. x ² + 5 x + 6 = 0 ( x + 3) ( x + 2) = 0 x = -3, -2 II. y²+ 7 y + 12= 0 ( y + 4) ( y + 3) = 0 y = -3, -4 Hence, x ≥ y
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