Question Number 203000 by sonukgindia last updated on 07/Jan/24
Answered by SEKRET last updated on 07/Jan/24
$$\mathrm{3};\mathrm{6} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Jan/24
$${xy}^{\mathrm{2}} +{xy}+{x}=\mathrm{129} \\ $$$${x}\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)=\mathrm{129}=\mathrm{1}×\mathrm{3}×\mathrm{43} \\ $$$$\mathrm{0}<{x}<{y}\:\wedge\:{x},{y}\in\mathbb{N}\:\wedge\:{y}>\mathrm{1} \\ $$$${y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{3}\:\mid\:{y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{43} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{2}=\mathrm{0}\:\mid\:{y}^{\mathrm{2}} +{y}−\mathrm{42}=\mathrm{0} \\ $$$$\left({y}+\mathrm{2}\right)\left({y}−\mathrm{1}\right)=\mathrm{0}\:\mid\:\left({y}+\mathrm{7}\right)\left({y}−\mathrm{6}\right)=\mathrm{0} \\ $$$${y}=\overset{×} {\mathrm{1}},\overset{×} {−\mathrm{2}}\:\mid\:{y}=−\overset{×} {\mathrm{7}}\:,\:{y}=\mathrm{6}\:\checkmark \\ $$$${x}\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)=\mathrm{129} \\ $$$$\Rightarrow{x}\left(\mathrm{6}^{\mathrm{2}} +\mathrm{6}+\mathrm{1}\right)=\mathrm{129} \\ $$$$\Rightarrow{x}=\frac{\mathrm{129}}{\mathrm{43}}=\mathrm{3} \\ $$$$\left({x},{y}\right)=\left(\mathrm{3},\mathrm{6}\right) \\ $$$$\mathrm{129}_{\mathrm{10}} =\mathrm{333}_{\mathrm{6}} \\ $$