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(D) For equation $I$: $x^{2}+13x+40=0$Factorizing the quadratic equation: $x^{2}+8x+5x+40=0$$x(x+8)+5(x+8)=0$$(x+5)(x+8)=0$Therefore,the roots are $x

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if $x \le y$

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(D) For equation $I$: $x^{2}+13x+40=0$Factorizing the quadratic equation: $x^{2}+8x+5x+40=0$$x(x+8)+5(x+8)=0$$(x+5)(x+8)=0$Therefore,the roots are $x = -5$ and $x = -8$.For equation $II$: $y^{2}+7y+10=0$Factorizing the quadratic equation: $y^{2}+5y+2y+10=0$$y(y+5)+2(y+5)=0$$(y+2)(y+5)=0$Therefore,the roots are $y = -2$ and $y = -5$.Comparing the values:When $x = -5$,$y$ can be $-2$ or $-5$. Here $x \le y$.When $x = -8$,$y$ can be $-2$ or $-5$. Here $x Combining these,we get $x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+13x+40=0$
$II.$ $y^{2}+7y+10=0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}+13x+40=0$$II.$ $y^{2}+7y+10=0$