$I . (6 x^{2} + 17) - (3 x^{2} + 20) = 0$
$I I . (5 y^{2} - 12) - (9 y^{2} - 16) = 0$

x < y

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x \geq y

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x \leq y

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x > y

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x=y or relationship cannot be established

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Solution

The correct option is E x=y or relationship cannot be established
Equation I
$(6 x^{2} + 17) - (3 x^{2} + 20) = 0$
$\Rightarrow (6 x^{2} - 3 x^{2}) + (17 - 20) = 0$
$\Rightarrow 3 x^{2} - 3 = 0$
$\Rightarrow 3 x^{2} = 3$
$∴ x = \pm 1$ .
Equation II
$(5 y^{2} - 12) - (9 y^{2} - 16) = 0$
$\Rightarrow (5 y^{2} - 9 y^{2}) + (- 12 + 16) = 0$
$\Rightarrow - 4 y^{2} + 4 = 0$
$∴ y = \pm 1$
Hence, x = y

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