$I . x^{2} + 20 = 9 x$
$I I . y^{2} + 42 = 13 y$

\geq

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x > y

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\leq

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x < y

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x = y or no relation can be established between x and y

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Solution

The correct option is D x < y
Equation I
$x^{2} - 9 x + 20 = 0$
$\Rightarrow x^{2} - 5 x - 4 x + 20 = 0$
$\Rightarrow x^{2} - 5 x - 4 x + 20 = 0$
$\Rightarrow x (x - 5) - 4 (x - 5) = 0$
$\Rightarrow (x - 4) (x - 5) = 0$
$∴ x = 4 o r 5$
Equation II
$y^{2} - 13 y + 42 = 0$
$\Rightarrow y^{2} - 7 y - 6 y + 42 = 0$
$\Rightarrow y (y - 7) - 6 (y - 7) = 0$
$\Rightarrow (y - 6) (y - 7) = 0$
$∴ y = 6 o r 7$
Hence, x < y

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x^{2} + 20 = 9 x

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