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(D) For equation $I$: $2x^2 - 11x + 15 = 0$$2x^2 - 6x - 5x + 15 = 0$$2x(x - 3) - 5(x - 3) = 0$$(2x - 5)(x - 3) = 0$So,$x = 2.5$ or $x = 3$.For equatio

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If relationship between $x$ and $y$ cannot be determined

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(D) For equation $I$: $2x^2 - 11x + 15 = 0$$2x^2 - 6x - 5x + 15 = 0$$2x(x - 3) - 5(x - 3) = 0$$(2x - 5)(x - 3) = 0$So,$x = 2.5$ or $x = 3$.For equation $II$: $2y^2 - 11y + 14 = 0$Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$y = \frac{11 \pm \sqrt{121 - 4(2)(14)}}{2(2)}$$y = \frac{11 \pm \sqrt{121 - 112}}{4}$$y = \frac{11 \pm \sqrt{9}}{4}$$y = \frac{11 \pm 3}{4}$So,$y = \frac{14}{4} = 3.5$ or $y = \frac{8}{4} = 2$.Comparing the values:If $x = 2.5$,then $x  y$ (since $y = 2$).Since the relationship varies,the relationship between $x$ and $y$ cannot be determined.

In these questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and mark the appropriate option.
$I.$ $2x^2 - 11x + 15 = 0$
$II.$ $2y^2 - 11y + 14 = 0$

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In these questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and mark the appropriate option.$I.$ $2x^2 - 11x + 15 = 0$$II.$ $2y^2 - 11y + 14 = 0$