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Question Number 171836 by Mikenice last updated on 21/Jun/22
solve:  x+3y=1  xy=y^2 −3
$${solve}: \\ $$$${x}+\mathrm{3}{y}=\mathrm{1} \\ $$$${xy}={y}^{\mathrm{2}} −\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jun/22
x+3y=1⇒x=1−3y  xy=y^2 −3⇒y(1−3y)=y^2 −3  ⇒y−3y^2 =y^2 −3  4y^2 −y−3=0  (y−1)(4y+3)=0  y=1 ∣  y=−(3/4)  •x=1−3y=1−3(1)=−2  • x=1−3(−(3/4))=((13)/4)  (x,y)={(((13)/4),−(3/4)),(−2,1)}
$${x}+\mathrm{3}{y}=\mathrm{1}\Rightarrow{x}=\mathrm{1}−\mathrm{3}{y} \\ $$$${xy}={y}^{\mathrm{2}} −\mathrm{3}\Rightarrow{y}\left(\mathrm{1}−\mathrm{3}{y}\right)={y}^{\mathrm{2}} −\mathrm{3} \\ $$$$\Rightarrow{y}−\mathrm{3}{y}^{\mathrm{2}} ={y}^{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} −{y}−\mathrm{3}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{4}{y}+\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\:\mid\:\:{y}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\bullet{x}=\mathrm{1}−\mathrm{3}{y}=\mathrm{1}−\mathrm{3}\left(\mathrm{1}\right)=−\mathrm{2} \\ $$$$\bullet\:{x}=\mathrm{1}−\mathrm{3}\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\left({x},{y}\right)=\left\{\left(\frac{\mathrm{13}}{\mathrm{4}},−\frac{\mathrm{3}}{\mathrm{4}}\right),\left(−\mathrm{2},\mathrm{1}\right)\right\} \\ $$

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