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(A) For equation $I: 20 x^{2}-67 x+56=0$We factorize the quadratic equation: $20 x^{2}-35 x-32 x+56=0$$5 x(4 x-7)-8(4 x-7)=0$$(5 x-8)(4 x-7)=0$So,$x =

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if $x > y$

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(A) For equation $I: 20 x^{2}-67 x+56=0$We factorize the quadratic equation: $20 x^{2}-35 x-32 x+56=0$$5 x(4 x-7)-8(4 x-7)=0$$(5 x-8)(4 x-7)=0$So,$x = \frac{8}{5} = 1.6$ or $x = \frac{7}{4} = 1.75$For equation $II: 56 y^{2}-67 y+20=0$We factorize the quadratic equation: $56 y^{2}-32 y-35 y+20=0$$8 y(7 y-4)-5(7 y-4)=0$$(8 y-5)(7 y-4)=0$So,$y = \frac{5}{8} = 0.625$ or $y = \frac{4}{7} \approx 0.57$Comparing the values: $x_1 = 1.6, x_2 = 1.75$ and $y_1 = 0.625, y_2 = 0.57$Since both values of $x$ are greater than both values of $y$,we conclude $x > y$.

Solve the given two equations and choose the correct option.
$I. 20 x^{2}-67 x+56=0$
$II. 56 y^{2}-67 y+20=0$

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Solve the given two equations and choose the correct option.$I. 20 x^{2}-67 x+56=0$$II. 56 y^{2}-67 y+20=0$