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(D) For equation $I$: Let $t = \frac{x}{x-11}$. Then $t + \frac{1}{t} = 7 \Rightarrow t^2 - 7t + 1 = 0$. Solving for $t$,$t = \frac{7 \pm \sqrt{49-4}}

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

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(D) For equation $I$: Let $t = \frac{x}{x-11}$. Then $t + \frac{1}{t} = 7 \Rightarrow t^2 - 7t + 1 = 0$. Solving for $t$,$t = \frac{7 \pm \sqrt{49-4}}{2} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3\sqrt{5}}{2}$.Since $t = \frac{x}{x-11}$,we have $x = t(x-11) = tx - 11t \Rightarrow x(1-t) = -11t \Rightarrow x = \frac{11t}{t-1}$.Substituting $t_1 = \frac{7+3\sqrt{5}}{2} \approx 6.85$ and $t_2 = \frac{7-3\sqrt{5}}{2} \approx 0.146$,we get two values for $x$.For equation $II$: Let $u = \frac{4y}{4y-13}$. Then $u + \frac{1}{u} = 9 \Rightarrow u^2 - 9u + 1 = 0$. Solving for $u$,$u = \frac{9 \pm \sqrt{81-4}}{2} = \frac{9 \pm \sqrt{77}}{2}$.Since $u = \frac{4y}{4y-13}$,we have $4y = u(4y-13) = 4uy - 13u \Rightarrow 4y(1-u) = -13u \Rightarrow y = \frac{13u}{4(u-1)}$.Since both equations yield two real roots (one positive and one negative),and the coefficients lead to overlapping ranges for $x$ and $y$,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct option.
$I.$ $\frac{x}{x-11} + \frac{x-11}{x} = 7$
$II.$ $\frac{4y}{4y-13} + \frac{4y-13}{4y} = 9$

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Solve the given two equations and select the correct option.$I.$ $\frac{x}{x-11} + \frac{x-11}{x} = 7$$II.$ $\frac{4y}{4y-13} + \frac{4y-13}{4y} = 9$