In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:
I. $6 x^{2} + 13 x - 28 = 0$
II. $3 y^{2} - 25 y + 28 = 0$

if x > y

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if x < y

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if x

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if x

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if x = y or relationship between x and y cannot be determined

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Solution

The correct option is D if xy
$6 x^{2} + 13 x - 28 = 0 \Rightarrow (3 x - 4) (2 x + 7) = 0 \Rightarrow x = \frac{4}{3}, - \frac{7}{2} 3 y^{2} - 25 y + 28 = 0 \Rightarrow (3 y - 4) (y - 7) = 0 \Rightarrow y = \frac{4}{3}, 7 ∴ x \leq y$

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