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(B) For equation $I$: $x - 7\sqrt{x} + 12 = 0$. Let $\sqrt{x} = u$,then $u^2 - 7u + 12 = 0$.$(u - 3)(u - 4) = 0$,so $u = 3$ or $u = 4$.Thus,$\sqrt{x}

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if $x \ge y$

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(B) For equation $I$: $x - 7\sqrt{x} + 12 = 0$. Let $\sqrt{x} = u$,then $u^2 - 7u + 12 = 0$.$(u - 3)(u - 4) = 0$,so $u = 3$ or $u = 4$.Thus,$\sqrt{x} = 3 \Rightarrow x = 9$ and $\sqrt{x} = 4 \Rightarrow x = 16$.For equation $II$: $y - 5\sqrt{y} + 6 = 0$. Let $\sqrt{y} = v$,then $v^2 - 5v + 6 = 0$.$(v - 2)(v - 3) = 0$,so $v = 2$ or $v = 3$.Thus,$\sqrt{y} = 2 \Rightarrow y = 4$ and $\sqrt{y} = 3 \Rightarrow y = 9$.Comparing the values: $x \in \{9, 16\}$ and $y \in \{4, 9\}$.Since all values of $x$ are greater than or equal to the values of $y$ $(9 \ge 4, 9 \ge 9, 16 \ge 4, 16 \ge 9)$,we conclude $x \ge y$.

Solve the given two equations and select the correct answer from the given options.
$I. \quad x - 7\sqrt{x} + 12 = 0$
$II. \quad y - 5\sqrt{y} + 6 = 0$

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Solve the given two equations and select the correct answer from the given options.$I. \quad x - 7\sqrt{x} + 12 = 0$$II. \quad y - 5\sqrt{y} + 6 = 0$