I. 6y2 – 23y + 20 = 0
II. 4x2 – 24 x + 35 = 0
I. 6y2 – 23y + 20 = 0 6y2 – 8y – 15 y + 20 = 0 2y (3 y – 4) – 5 (3 y – 4) = 0 (2 y – 5) (3 y – 4) = 0 y = 5/2 ,4/3 II. 4x2 – 24 x + 35 = 0 4x2 – 14 x – 10 x + 35 = 0 2 x(2 x – 7) – 5 (2 x – 7) = 0 (2 x – 5) (2 x – 7) = 0 x = 5/2 ,7/2 Hence, x≥y
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