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(D) Step $1$: Solve equation $I$.$\frac{15}{\sqrt{x}} - \frac{9}{\sqrt{x}} = \sqrt{x}$$\frac{6}{\sqrt{x}} = \sqrt{x}$$6 = \sqrt{x} \cdot \sqrt{x}$$x =

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

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(D) Step $1$: Solve equation $I$.$\frac{15}{\sqrt{x}} - \frac{9}{\sqrt{x}} = \sqrt{x}$$\frac{6}{\sqrt{x}} = \sqrt{x}$$6 = \sqrt{x} \cdot \sqrt{x}$$x = 6$Step $2$: Solve equation $II$.$y^{10} - (36)^{5} = 0$$y^{10} = (36)^{5}$Since $36 = 6^2$,we have $y^{10} = (6^2)^5 = 6^{10}$.Taking the $10^{th}$ root on both sides,$y = \pm 6$.Step $3$: Compare $x$ and $y$.$x = 6$ and $y = 6$ or $y = -6$.If $y = 6$,then $x = y$.If $y = -6$,then $x > y$.Since both conditions are possible,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{15}{\sqrt{x}} - \frac{9}{\sqrt{x}} = (x)^{1/2}$
$II.$ $y^{10} - (36)^{5} = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $\frac{15}{\sqrt{x}} - \frac{9}{\sqrt{x}} = (x)^{1/2}$$II.$ $y^{10} - (36)^{5} = 0$