Directions: In the following question, two equations (I) and (II) are given. You have to solve both the equations and give answer according to the codes given below: (a) if x > y (b) if x > y (c) if x y (e) if x = y, or the relation between x and y cannot be established I. 16x2 + 6x − 1 = 0 II. 32y2 − 28y + 3 = 0

Solving equation I, 16x^2 + 6x - 1 = 0, by splitting the middle term gives 16x^2 + 8x - 2x - 1 = 0, which factors into (4x - 1)(4x + 2) = 0, yielding x = 1/4 and x = -1/2. Solving equation II, 32y^2 - 28y + 3 = 0, gives 32y^2 - 24y - 4y + 3 = 0, which factors into (4y - 3)(8y - 1) = 0, yielding y = 3/4 and y = 1/8. Comparing the values, x = 1/4 is less than y = 3/4 but greater than y = 1/8, so no clear relationship can be established.

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