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(A) For equation $I$: $x^{2}-7 \sqrt{3} x-5 \sqrt{5} x+35 \sqrt{15}=0$$x(x-7 \sqrt{3})-5 \sqrt{5}(x-7 \sqrt{3})=0$$(x-5 \sqrt{5})(x-7 \sqrt{3})=0$So,$

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if $x > y$

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(A) For equation $I$: $x^{2}-7 \sqrt{3} x-5 \sqrt{5} x+35 \sqrt{15}=0$$x(x-7 \sqrt{3})-5 \sqrt{5}(x-7 \sqrt{3})=0$$(x-5 \sqrt{5})(x-7 \sqrt{3})=0$So,$x = 5 \sqrt{5}$ or $x = 7 \sqrt{3}$.For equation $II$: $y^{2}-5 \sqrt{5} y+30=0$$y^{2}-3 \sqrt{5} y -2 \sqrt{5} y +30=0$$y(y-3 \sqrt{5})-2 \sqrt{5}(y-3 \sqrt{5})=0$$(y-2 \sqrt{5})(y-3 \sqrt{5})=0$So,$y = 2 \sqrt{5}$ or $y = 3 \sqrt{5}$.Comparing the values:$5 \sqrt{5} \approx 5 	imes 2.236 = 11.18$$7 \sqrt{3} \approx 7 	imes 1.732 = 12.124$$2 \sqrt{5} \approx 2 	imes 2.236 = 4.472$$3 \sqrt{5} \approx 3 	imes 2.236 = 6.708$Since all values of $x$ are greater than all values of $y$,we conclude $x > y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-7 \sqrt{3} x+35 \sqrt{15}=5 \sqrt{5} x$
$II.$ $y^{2}-5 \sqrt{5} y+30=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-7 \sqrt{3} x+35 \sqrt{15}=5 \sqrt{5} x$$II.$ $y^{2}-5 \sqrt{5} y+30=0$