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(D) For equation $I$: $x^{2} - 25 = 0 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$.So,$x = 5$ or $x = -5$.For equation $II$: $y^{2} - 9y + 20 = 0$.Fa

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if $x = y$ or relationship between $x$ and $y$ cannot be established.

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(D) For equation $I$: $x^{2} - 25 = 0 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$.So,$x = 5$ or $x = -5$.For equation $II$: $y^{2} - 9y + 20 = 0$.Factoring the quadratic equation: $y^{2} - 5y - 4y + 20 = 0 \Rightarrow y(y - 5) - 4(y - 5) = 0 \Rightarrow (y - 5)(y - 4) = 0$.So,$y = 5$ or $y = 4$.Comparing the values:If $x = 5$,then $x = y$ (when $y = 5$) and $x > y$ (when $y = 4$).If $x = -5$,then $x Since we get different results ($x = y$,$x > y$,and $x < y$) depending on the values chosen,a unique relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $y^{2}-9y+20=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-25=0$$II.$ $y^{2}-9y+20=0$