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(B) For equation $I$: $676 x^{2} - 1 = 0$$676 x^{2} = 1$$x^{2} = \frac{1}{676}$$x = \pm \frac{1}{26}$So,$x = \frac{1}{26} \approx 0.0384$ or $x = -\fr

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if $x < y$

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(B) For equation $I$: $676 x^{2} - 1 = 0$$676 x^{2} = 1$$x^{2} = \frac{1}{676}$$x = \pm \frac{1}{26}$So,$x = \frac{1}{26} \approx 0.0384$ or $x = -\frac{1}{26} \approx -0.0384$.For equation $II$: $y = \frac{1}{\sqrt[3]{13824}}$Since $24^{3} = 13824$,we have $\sqrt[3]{13824} = 24$.Thus,$y = \frac{1}{24} \approx 0.0416$.Comparing the values:$x = 0.0384$ or $-0.0384$ and $y = 0.0416$.In both cases,$x < y$.

Solve the given two equations and select the correct option.
$I.$ $676 x^{2}-1=0$
$II.$ $y=\frac{1}{\sqrt[3]{13824}}$

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Solve the given two equations and select the correct option.$I.$ $676 x^{2}-1=0$$II.$ $y=\frac{1}{\sqrt[3]{13824}}$