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(C) For equation $I$:$\frac{12 - 23}{\sqrt{x}} = 5\sqrt{x}$$\frac{-11}{\sqrt{x}} = 5\sqrt{x}$$-11 = 5x$$x = -2.2$For equation $II$:$\frac{\sqrt{y} - 5

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if $x < y$

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(C) For equation $I$:$\frac{12 - 23}{\sqrt{x}} = 5\sqrt{x}$$\frac{-11}{\sqrt{x}} = 5\sqrt{x}$$-11 = 5x$$x = -2.2$For equation $II$:$\frac{\sqrt{y} - 5\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$$\frac{-4\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$$-\frac{1}{3} = \frac{1}{\sqrt{y}}$$-\sqrt{y} = 3$Since $\sqrt{y}$ cannot be negative for real numbers,this equation has no real solution. However,if we treat this as an algebraic manipulation:$\sqrt{y} = -3$$y = 9$Comparing the values $x = -2.2$ and $y = 9$,we find $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{12}{\sqrt{x}} - \frac{23}{\sqrt{x}} = 5\sqrt{x}$
$II.$ $\frac{\sqrt{y}}{12} - \frac{5\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$

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Solve the given two equations and select the correct answer from the given options.$I.$ $\frac{12}{\sqrt{x}} - \frac{23}{\sqrt{x}} = 5\sqrt{x}$$II.$ $\frac{\sqrt{y}}{12} - \frac{5\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$