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(D) For equation $I$: $x^{2}-25=0 \Rightarrow x^{2}=25 \Rightarrow x = 5, -5$.For equation $II$: $4y^{2}-24y+35=0$.Using the quadratic formula or fact

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

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(D) For equation $I$: $x^{2}-25=0 \Rightarrow x^{2}=25 \Rightarrow x = 5, -5$.For equation $II$: $4y^{2}-24y+35=0$.Using the quadratic formula or factorization: $4y^{2}-14y-10y+35=0$.$2y(2y-7)-5(2y-7)=0 \Rightarrow (2y-5)(2y-7)=0$.Thus,$y = 2.5$ or $y = 3.5$.Comparing the values:If $x = 5$,then $x > 2.5$ and $x > 3.5$ $(x > y)$.If $x = -5$,then $x Since we get different relationships ($x > y$ and $x < y$),the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-25=0$
$II.$ $4y^{2}=24y-35$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-25=0$$II.$ $4y^{2}=24y-35$