Question
I. 2 x ² + x  – 1 =
0                 II. 2 y ² - 3 y + 1 = 0Solution
I. 2 x ² + x – 1 = 0 (x – 1) (2 x + 1) = 0 x = 1, - 1/2 II. 2 y ² - 3 y + 1 = 0 (2 y – 1) ( y – 1) = 0 y = 1, 1/2 Hence,  x ≤ y
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