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(D) For 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$For equation $I

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

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(D) For equation $I$:$\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$$\frac{6}{\sqrt{x}}=\sqrt{x}$$6 = \sqrt{x} \cdot \sqrt{x}$$x = 6$For equation $II$:$y^{10} - (36)^{5} = 0$$y^{10} = (6^2)^5$$y^{10} = 6^{10}$Since the exponent is even,$y = 6$ or $y = -6$.Comparing $x$ and $y$:If $x = 6$ and $y = 6$,then $x = y$.If $x = 6$ and $y = -6$,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.
$I.$ $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=(x)^{\frac{1}{2}}$
$II.$ $y^{10}-(36)^{5}=0$

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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)^{\frac{1}{2}}$$II.$ $y^{10}-(36)^{5}=0$