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(A) For equation $I$: $4x^2 - 43x + 105 = 0$We factorize the quadratic equation: $4x^2 - 28x - 15x + 105 = 0$$4x(x - 7) - 15(x - 7) = 0$$(4x - 15)(x -

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if $x > y$

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(A) For equation $I$: $4x^2 - 43x + 105 = 0$We factorize the quadratic equation: $4x^2 - 28x - 15x + 105 = 0$$4x(x - 7) - 15(x - 7) = 0$$(4x - 15)(x - 7) = 0$So,$x_1 = \frac{15}{4} = 3.75$ and $x_2 = 7$.For equation $II$: $7y^2 - 29y + 30 = 0$We factorize the quadratic equation: $7y^2 - 14y - 15y + 30 = 0$$7y(y - 2) - 15(y - 2) = 0$$(7y - 15)(y - 2) = 0$So,$y_1 = \frac{15}{7} \approx 2.14$ and $y_2 = 2$.Comparing the values:$x_1 = 3.75$ and $x_2 = 7$$y_1 \approx 2.14$ and $y_2 = 2$Since all values of $x$ are greater than all values of $y$ $(3.75 > 2.14, 3.75 > 2, 7 > 2.14, 7 > 2)$,we conclude that $x > y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $4x^2 - 43x + 105 = 0$
$II.$ $7y^2 - 29y + 30 = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $4x^2 - 43x + 105 = 0$$II.$ $7y^2 - 29y + 30 = 0$