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Question Number 115909 by Eric002 last updated on 29/Sep/20
solve the system of equations x+((3x−y)/(x^2 +y^2 ))=3 , y−((x+3y)/(x^2 +y^2 ))=0
$${solve}\:{the}\:{system}\:{of}\:{equations} \\ $$$${x}+\frac{\mathrm{3}{x}−{y}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{3}\:,\:{y}−\frac{{x}+\mathrm{3}{y}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{0} \\ $$
Answered by MJS_new last updated on 29/Sep/20
let y=px (1) x+((3−p)/((1+p^2 )x))=3 (2) px−((1+3p)/((1+p^2 )x))=0 (1) x^2 =3x+((p−3)/(1+p^2 )) (2) x^2 =((3p+1)/((1+p^2 )p)) ⇒ 3x=((3p+1)/((1+p^2 )p))−((p−3)/(1+p^2 )) x=−((p^2 −6p−1)/(3(1+p^2 )p)) (1) x^2 −3x−((p−3)/(1+p^2 ))=0 ⇒ p^4 +((21)/(26))p^3 −(7/(26))p^2 −(3/(26))p−(1/(26))=0 (p+1)(p−(1/2))(p^2 +(4/(13))p+(1/(13)))=0 p_1 =−1 p_2 =(1/2) p_(3, 4) =−(2/(13))±(3/(13))i ⇒ x_1 =1∧y_1 =−1 x_2 =2∧y_2 =1 x_3 =(3/2)−i∧y_3 =(1/2)i x_4 =(3/2)+i∧y_4 =−(1/2)i
$$\mathrm{let}\:{y}={px} \\ $$$$\left(\mathrm{1}\right)\:{x}+\frac{\mathrm{3}−{p}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right){x}}=\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:{px}−\frac{\mathrm{1}+\mathrm{3}{p}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right){x}}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} =\mathrm{3}{x}+\frac{{p}−\mathrm{3}}{\mathrm{1}+{p}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} =\frac{\mathrm{3}{p}+\mathrm{1}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right){p}} \\ $$$$\Rightarrow \\ $$$$\mathrm{3}{x}=\frac{\mathrm{3}{p}+\mathrm{1}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right){p}}−\frac{{p}−\mathrm{3}}{\mathrm{1}+{p}^{\mathrm{2}} } \\ $$$${x}=−\frac{{p}^{\mathrm{2}} −\mathrm{6}{p}−\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{p}^{\mathrm{2}} \right){p}} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\frac{{p}−\mathrm{3}}{\mathrm{1}+{p}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{4}} +\frac{\mathrm{21}}{\mathrm{26}}{p}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{26}}{p}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{26}}{p}−\frac{\mathrm{1}}{\mathrm{26}}=\mathrm{0} \\ $$$$\left({p}+\mathrm{1}\right)\left({p}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({p}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{13}}{p}+\frac{\mathrm{1}}{\mathrm{13}}\right)=\mathrm{0} \\ $$$${p}_{\mathrm{1}} =−\mathrm{1} \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{2}}{\mathrm{13}}\pm\frac{\mathrm{3}}{\mathrm{13}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\wedge{y}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}\wedge{y}_{\mathrm{2}} =\mathrm{1} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{i}\wedge{y}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$${x}_{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{i}\wedge{y}_{\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$

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