I. (x13/5 ÷7) = 5488 ÷ x7/5
II. (y2/3 × y2/3 ) ÷ √4 = (343y)1/3
I. (x(13/5) ÷7) = 5488 ÷ x(7/5) x(13/5 + 7/5) = 5488 ×7 x4 = 38416 x = ± 14 II. {(y)(2/3) × y(2/3)} ÷ √4 = (343y)(1/3) y(4/3) ÷ y(1/3) = 2 × 7 y=14 Hence, y ≥ x
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