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(D) Step $1$: Solve equation $I$: $x^{2} = 16$. Taking the square root on both sides,we get $x = 4$ or $x = -4$.Step $2$: Solve equation $II$: $2y^{2}

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If $x \leq y$

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(D) Step $1$: Solve equation $I$: $x^{2} = 16$. Taking the square root on both sides,we get $x = 4$ or $x = -4$.Step $2$: Solve equation $II$: $2y^{2} - 17y + 36 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,where $a = 2, b = -17, c = 36$.$y = \frac{17 \pm \sqrt{(-17)^{2} - 4(2)(36)}}{2(2)}$$y = \frac{17 \pm \sqrt{289 - 288}}{4}$$y = \frac{17 \pm 1}{4}$So,$y = \frac{18}{4} = 4.5$ or $y = \frac{16}{4} = 4$.Step $3$: Compare the values of $x$ and $y$:For $x = 4$,$y$ can be $4$ or $4.5$. Here $x \leq y$.For $x = -4$,$y$ can be $4$ or $4.5$. Here $x Combining these,we get $x \leq y$.

In these questions,two equations numbered $I$ and $II$ are given. You have to solve both the equations and mark the appropriate option.
$I$. $x^{2} = 16$
$II$. $2y^{2} - 17y + 36 = 0$

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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.$I$. $x^{2} = 16$$II$. $2y^{2} - 17y + 36 = 0$