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(D) For equation $I: 3x^2 - 4x - 32 = 0$Using the quadratic formula or factorization: $3x^2 - 12x + 8x - 32 = 0$$3x(x - 4) + 8(x - 4) = 0$$(3x + 8)(x

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if $x \le y$

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(D) For equation $I: 3x^2 - 4x - 32 = 0$Using the quadratic formula or factorization: $3x^2 - 12x + 8x - 32 = 0$$3x(x - 4) + 8(x - 4) = 0$$(3x + 8)(x - 4) = 0$So,$x = 4$ or $x = -8/3 \approx -2.67$.For equation $II: 2y^2 - 17y + 36 = 0$Factorizing: $2y^2 - 9y - 8y + 36 = 0$$y(2y - 9) - 4(2y - 9) = 0$$(y - 4)(2y - 9) = 0$So,$y = 4$ or $y = 9/2 = 4.5$.Comparing the values:$x = \{4, -2.67\}$$y = \{4, 4.5\}$Since $4 \le 4$,$4 \le 4.5$,$-2.67 \le 4$,and $-2.67 \le 4.5$,we conclude that $x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

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Solve the given two equations and select the correct answer from the given options.$I. 3x^2 - 4x - 32 = 0$$II. 2y^2 - 17y + 36 = 0$