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(A) Step $1$: Solve the first equation $x^{2}-4=0$.$x^{2} = 4$$x = \pm 2$So,$x = 2$ or $x = -2$.Step $2$: Solve the second equation $y^{2}+6y+9=0$.Thi

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if $x > y$

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(A) Step $1$: Solve the first equation $x^{2}-4=0$.$x^{2} = 4$$x = \pm 2$So,$x = 2$ or $x = -2$.Step $2$: Solve the second equation $y^{2}+6y+9=0$.This is a perfect square trinomial: $(y+3)^{2} = 0$.$y+3 = 0$$y = -3$.Step $3$: Compare the values of $x$ and $y$.We have $x = 2, -2$ and $y = -3$.Comparing $x = 2$ with $y = -3$,we get $2 > -3$ (i.e.,$x > y$).Comparing $x = -2$ with $y = -3$,we get $-2 > -3$ (i.e.,$x > y$).In both cases,$x > y$.Therefore,the correct option is $A$.

Solve the given two equations and select the correct option.
$I.$ $x^{2}-4=0$
$II.$ $y^{2}+6y+9=0$

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Solve the given two equations and select the correct option.$I.$ $x^{2}-4=0$$II.$ $y^{2}+6y+9=0$