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(B) For equation $I$: $2x^2 - 4x - \sqrt{13}x + 2\sqrt{13} = 0$$2x(x - 2) - \sqrt{13}(x - 2) = 0$$(x - 2)(2x - \sqrt{13}) = 0$Thus,$x = 2$ or $x = \fr

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if $x \ge y$

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(B) For equation $I$: $2x^2 - 4x - \sqrt{13}x + 2\sqrt{13} = 0$$2x(x - 2) - \sqrt{13}(x - 2) = 0$$(x - 2)(2x - \sqrt{13}) = 0$Thus,$x = 2$ or $x = \frac{\sqrt{13}}{2} \approx \frac{3.605}{2} \approx 1.8025$.For equation $II$: $10y^2 - 18y - 5\sqrt{13}y + 9\sqrt{13} = 0$$2y(5y - 9) - \sqrt{13}(5y - 9) = 0$$(2y - \sqrt{13})(5y - 9) = 0$Thus,$y = \frac{\sqrt{13}}{2} \approx 1.8025$ or $y = \frac{9}{5} = 1.8$.Comparing the values:$x_1 = 2, x_2 \approx 1.8025$$y_1 \approx 1.8025, y_2 = 1.8$Since $2 > 1.8025$ and $2 > 1.8$,and $1.8025 > 1.8025$ (equal) and $1.8025 > 1.8$,we observe that $x \ge y$.

Solve the given two equations and choose the correct option.
$I.$ $2x^2 - (4 + \sqrt{13})x + 2\sqrt{13} = 0$
$II.$ $10y^2 - (18 + 5\sqrt{13})y + 9\sqrt{13} = 0$

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Solve the given two equations and choose the correct option.$I.$ $2x^2 - (4 + \sqrt{13})x + 2\sqrt{13} = 0$$II.$ $10y^2 - (18 + 5\sqrt{13})y + 9\sqrt{13} = 0$