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(D) For equation $I$: $x^{2}-26x+168=0$We need two numbers whose sum is $26$ and product is $168$. These numbers are $12$ and $14$.$x^{2}-12x-14x+168=

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

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(D) For equation $I$: $x^{2}-26x+168=0$We need two numbers whose sum is $26$ and product is $168$. These numbers are $12$ and $14$.$x^{2}-12x-14x+168=0$$x(x-12)-14(x-12)=0$$(x-12)(x-14)=0$Therefore,$x = 12$ or $x = 14$.For equation $II$: $y^{2}-25y+156=0$We need two numbers whose sum is $25$ and product is $156$. These numbers are $12$ and $13$.$y^{2}-12y-13y+156=0$$y(y-12)-13(y-12)=0$$(y-12)(y-13)=0$Therefore,$y = 12$ or $y = 13$.Comparing the values:If $x=12$ and $y=12$,then $x=y$.If $x=12$ and $y=13$,then $xIf $x=14$ and $y=12$,then $x>y$.Since we get different results ($x=y$,$xy$),the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-26x+168=0$
$II.$ $y^{2}-25y+156=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-26x+168=0$$II.$ $y^{2}-25y+156=0$