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(D) Step $1$: Solve equation $I$.$\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$$\frac{9 + 19}{\sqrt{x}} = \sqrt{x}$$\frac{28}{\sqrt{x}} = \sqrt

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if $x = y$ or the relationship cannot be established.

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(D) Step $1$: Solve equation $I$.$\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$$\frac{9 + 19}{\sqrt{x}} = \sqrt{x}$$\frac{28}{\sqrt{x}} = \sqrt{x}$$28 = \sqrt{x} 	imes \sqrt{x}$$x = 28$Step $2$: Solve equation $II$.$y^{5} - \frac{(28)^{1/2}}{\sqrt{y}} = 0$$y^{5} = \frac{(28)^{1/2}}{y^{1/2}}$$y^{5} 	imes y^{1/2} = (28)^{1/2}$$y^{11/2} = (28)^{1/2}$Since the exponents are the same,the bases must be equal.$y = 28$Conclusion:Since $x = 28$ and $y = 28$,we have $x = y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$
$II.$ $y^{5} - \frac{(28)^{1/2}}{\sqrt{y}} = 0$

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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.$ $y^{5} - \frac{(28)^{1/2}}{\sqrt{y}} = 0$