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(D) Let $t_1 = \frac{x}{x+7}$. Then the equation $I$ becomes $t_1 + \frac{1}{t_1} = 12$,which implies $t_1^2 - 12t_1 + 1 = 0$. Solving for $t_1$ using

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

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(D) Let $t_1 = \frac{x}{x+7}$. Then the equation $I$ becomes $t_1 + \frac{1}{t_1} = 12$,which implies $t_1^2 - 12t_1 + 1 = 0$. Solving for $t_1$ using the quadratic formula,$t_1 = \frac{12 \pm \sqrt{144 - 4}}{2} = 6 \pm \sqrt{35}$.Since $t_1 = \frac{x}{x+7}$,we have $x = t_1(x+7) \Rightarrow x(1-t_1) = 7t_1 \Rightarrow x = \frac{7t_1}{1-t_1}$.For $t_1 = 6 + \sqrt{35} \approx 11.916$,$x = \frac{7(11.916)}{1-11.916} \approx -7.63$.For $t_1 = 6 - \sqrt{35} \approx 0.084$,$x = \frac{7(0.084)}{1-0.084} \approx 0.64$.Similarly for equation $II$,let $t_2 = \frac{y}{y+8}$. Then $t_2 + \frac{1}{t_2} = 16$,which implies $t_2^2 - 16t_2 + 1 = 0$. Solving for $t_2$,$t_2 = \frac{16 \pm \sqrt{256 - 4}}{2} = 8 \pm \sqrt{63} \approx 8 \pm 7.937$.$t_2 \approx 15.937$ or $t_2 \approx 0.063$.Using $y = \frac{8t_2}{1-t_2}$:For $t_2 \approx 15.937$,$y = \frac{8(15.937)}{1-15.937} \approx -8.53$.For $t_2 \approx 0.063$,$y = \frac{8(0.063)}{1-0.063} \approx 0.54$.Comparing the values of $x \in \{-7.63, 0.64\}$ and $y \in \{-8.53, 0.54\}$,we see that $x$ can be greater than,less than,or equal to $y$ depending on the roots chosen. Thus,the relationship cannot be established.

Solve the given two equations and provide the correct answer from the given options.
$I.$ $\frac{x}{x+7} + \frac{x+7}{x} = 12$
$II.$ $\frac{y}{y+8} + \frac{y+8}{y} = 16$

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Solve the given two equations and provide the correct answer from the given options.$I.$ $\frac{x}{x+7} + \frac{x+7}{x} = 12$$II.$ $\frac{y}{y+8} + \frac{y+8}{y} = 16$