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(D) For equation $I$:$\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$$\frac{9+19}{\sqrt{x}} = \sqrt{x}$$28 = \sqrt{x} 	imes \sqrt{x}$$x = 28$For

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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{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$$\frac{9+19}{\sqrt{x}} = \sqrt{x}$$28 = \sqrt{x} 	imes \sqrt{x}$$x = 28$For equation $II$:$y^{5} = \frac{(28)^{11/2}}{\sqrt{y}}$$y^{5} 	imes y^{1/2} = (28)^{11/2}$$y^{5 + 1/2} = (28)^{11/2}$$y^{11/2} = (28)^{11/2}$$y = 28$Comparing the values,$x = 28$ and $y = 28$,therefore $x = y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$
$II.$ $y^{5} - \frac{(2 \times 14)^{11/2}}{\sqrt{y}} = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$$II.$ $y^{5} - \frac{(2 \times 14)^{11/2}}{\sqrt{y}} = 0$