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(B) For equation $I$: $x^{2}-9x+20=0$.We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.So,$(x-5)(x-4)=0$,which gives $x

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if $x < y$

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(B) For equation $I$: $x^{2}-9x+20=0$.We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.So,$(x-5)(x-4)=0$,which gives $x = 5$ or $x = 4$.For equation $II$: $y^{2}-13y+42=0$.We need two numbers whose product is $42$ and sum is $13$. These are $7$ and $6$.So,$(y-7)(y-6)=0$,which gives $y = 7$ or $y = 6$.Comparing the values:If $x=4$,then $y=7$ $(x If $x=5$,then $y=7$ $(x In all cases,$x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+20=9x$
$II.$ $y^{2}+42=13y$

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Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}+20=9x$$II.$ $y^{2}+42=13y$