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(A) For equation $I: 6x^{2} + 29x + 35 = 0$We factorize the quadratic equation: $6x^{2} + 14x + 15x + 35 = 0$$2x(3x + 7) + 5(3x + 7) = 0$$(2x + 5)(3x

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if $x > y$

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(A) For equation $I: 6x^{2} + 29x + 35 = 0$We factorize the quadratic equation: $6x^{2} + 14x + 15x + 35 = 0$$2x(3x + 7) + 5(3x + 7) = 0$$(2x + 5)(3x + 7) = 0$Thus,$x = -2.5$ or $x = -2.33$.For equation $II: 3y^{2} + 19y + 30 = 0$We factorize the quadratic equation: $3y^{2} + 9y + 10y + 30 = 0$$3y(y + 3) + 10(y + 3) = 0$$(3y + 10)(y + 3) = 0$Thus,$y = -3.33$ or $y = -3$.Comparing the values:Since $-2.5 > -3.33$,$-2.5 > -3$,$-2.33 > -3.33$,and $-2.33 > -3$,we conclude that $x > y$.

Solve the given two equations and select the correct answer from the given options:
$I. 6x^{2} + 29x + 35 = 0$
$II. 3y^{2} + 19y + 30 = 0$

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Solve the given two equations and select the correct answer from the given options:$I. 6x^{2} + 29x + 35 = 0$$II. 3y^{2} + 19y + 30 = 0$