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(C) For equation $I: x - 7\sqrt{3x} + 36 = 0$Let $\sqrt{x} = u$. Then $u^2 - 7\sqrt{3}u + 36 = 0$.Using the quadratic formula or factoring: $u^2 - 4\s

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if $x < y$

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(C) For equation $I: x - 7\sqrt{3x} + 36 = 0$Let $\sqrt{x} = u$. Then $u^2 - 7\sqrt{3}u + 36 = 0$.Using the quadratic formula or factoring: $u^2 - 4\sqrt{3}u - 3\sqrt{3}u + 36 = 0$$u(u - 4\sqrt{3}) - 3\sqrt{3}(u - 4\sqrt{3}) = 0$$(u - 3\sqrt{3})(u - 4\sqrt{3}) = 0$So,$u = 3\sqrt{3} = \sqrt{27}$ or $u = 4\sqrt{3} = \sqrt{48}$.Thus,$x = 27$ or $x = 48$.For equation $II: y - 12\sqrt{2y} + 70 = 0$Let $\sqrt{y} = v$. Then $v^2 - 12\sqrt{2}v + 70 = 0$.Factoring: $v^2 - 7\sqrt{2}v - 5\sqrt{2}v + 70 = 0$$v(v - 7\sqrt{2}) - 5\sqrt{2}(v - 7\sqrt{2}) = 0$$(v - 5\sqrt{2})(v - 7\sqrt{2}) = 0$So,$v = 5\sqrt{2} = \sqrt{50}$ or $v = 7\sqrt{2} = \sqrt{98}$.Thus,$y = 50$ or $y = 98$.Comparing the values: $x \in \{27, 48\}$ and $y \in \{50, 98\}$.In all cases,$x < y$.

Solve the given two equations and select the correct answer from the given options.
$I. x - 7\sqrt{3x} + 36 = 0$
$II. y - 12\sqrt{2y} + 70 = 0$

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Solve the given two equations and select the correct answer from the given options.$I. x - 7\sqrt{3x} + 36 = 0$$II. y - 12\sqrt{2y} + 70 = 0$