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(B) For equation $I$: Let $k = \frac{3x}{3x+7}$. Then $k - \frac{1}{k} = 14 \implies k^2 - 14k - 1 = 0$.Using the quadratic formula $k = \frac{14 \pm

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if $x < y$

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(B) For equation $I$: Let $k = \frac{3x}{3x+7}$. Then $k - \frac{1}{k} = 14 \implies k^2 - 14k - 1 = 0$.Using the quadratic formula $k = \frac{14 \pm \sqrt{196 + 4}}{2} = 7 \pm 5\sqrt{2}$.Since $k = \frac{3x}{3x+7}$,we solve for $x$. Both roots for $x$ will be negative because the constant term in the resulting quadratic $126x^2 + 336x + 49 = 0$ is positive and the middle term is positive.For equation $II$: Let $m = \frac{y}{18y-5}$. Then $m - \frac{1}{m} = 2 \implies m^2 - 2m - 1 = 0$.Using the quadratic formula $m = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.Solving for $y$,we find that the roots for $y$ are positive because the resulting quadratic $359y^2 - 190y + 25 = 0$ has a positive constant term and a negative middle term.Since all values of $x$ are negative and all values of $y$ are positive,it follows that $x < y$.

Solve the given two equations and choose the correct option.
$I.$ $\frac{3x}{3x+7} - \frac{3x+7}{3x} = 14$
$II.$ $\frac{y}{18y-5} - \frac{18y-5}{y} = 2$

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Solve the given two equations and choose the correct option.$I.$ $\frac{3x}{3x+7} - \frac{3x+7}{3x} = 14$$II.$ $\frac{y}{18y-5} - \frac{18y-5}{y} = 2$