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(B) Step $1$: Solve equation $I$.$\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$Multiply the entire equation by $x^2$ (assuming $x 
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if $x < y$

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(B) Step $1$: Solve equation $I$.$\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$Multiply the entire equation by $x^2$ (assuming $x 
eq 0$):$18 + 6x - 12 = 8$$6 + 6x = 8$$6x = 2$$x = \frac{2}{6} = \frac{1}{3} \approx 0.333$Step $2$: Solve equation $II$.$y^3 + 9.68 + 5.64 = 16.95$$y^3 + 15.32 = 16.95$$y^3 = 16.95 - 15.32$$y^3 = 1.63$Since $1^3 = 1$ and $2^3 = 8$,$y$ must be slightly greater than $1$ (specifically $y \approx 1.177$).Step $3$: Compare $x$ and $y$.$x \approx 0.333$ and $y \approx 1.177$.Therefore,$x < y$.

Solve the given two equations and select the correct option.
$I.$ $\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$
$II.$ $y^3 + 9.68 + 5.64 = 16.95$

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Solve the given two equations and select the correct option.$I.$ $\frac{18}{x^2} + \frac{6}{x} - \frac{12}{x^2} = \frac{8}{x^2}$$II.$ $y^3 + 9.68 + 5.64 = 16.95$