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(B) For equation $I$: $x^{2} - 13x + 42 = 0$.Splitting the middle term: $x^{2} - 7x - 6x + 42 = 0$.$x(x - 7) - 6(x - 7) = 0$.$(x - 6)(x - 7) = 0$.Thus

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if $x \ge y$

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(B) For equation $I$: $x^{2} - 13x + 42 = 0$.Splitting the middle term: $x^{2} - 7x - 6x + 42 = 0$.$x(x - 7) - 6(x - 7) = 0$.$(x - 6)(x - 7) = 0$.Thus,$x = 6$ or $x = 7$.For equation $II$: $y = \sqrt[4]{1296}$.Since $6^{4} = 6 	imes 6 	imes 6 	imes 6 = 1296$,we have $y = 6$.Comparing the values:When $x = 6$,$x = y$.When $x = 7$,$x > y$.Combining these,we get $x \ge y$.

Solve the given two equations and select the correct option.
$I.$ $x^{2} + 42 = 13x$
$II.$ $y = \sqrt[4]{1296}$

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Solve the given two equations and select the correct option.$I.$ $x^{2} + 42 = 13x$$II.$ $y = \sqrt[4]{1296}$