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(D) For equation $I$: $x^{2}-5x+6=0$Factorizing the quadratic equation: $x^{2}-3x-2x+6=0$$x(x-3)-2(x-3)=0$$(x-3)(x-2)=0$Thus,the roots are $x = 3$ and

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if $x \le y$

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(D) For equation $I$: $x^{2}-5x+6=0$Factorizing the quadratic equation: $x^{2}-3x-2x+6=0$$x(x-3)-2(x-3)=0$$(x-3)(x-2)=0$Thus,the roots are $x = 3$ and $x = 2$.For equation $II$: $2y^{2}-15y+27=0$Factorizing the quadratic equation: $2y^{2}-6y-9y+27=0$$2y(y-3)-9(y-3)=0$$(2y-9)(y-3)=0$Thus,the roots are $y = \frac{9}{2} = 4.5$ and $y = 3$.Comparing the values:When $x = 3$,$y$ can be $4.5$ or $3$. Here $x \le y$.When $x = 2$,$y$ can be $4.5$ or $3$. Here $x Combining these,we get $x \le y$.

Solve the given two equations and select the correct option.
$I.$ $x^{2}-5x+6=0$
$II.$ $2y^{2}-15y+27=0$

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Solve the given two equations and select the correct option.$I.$ $x^{2}-5x+6=0$$II.$ $2y^{2}-15y+27=0$