Question-205211

Question Number 205211 by cortano12 last updated on 13/Mar/24

Answered by Berbere last updated on 13/Mar/24

{ ((5x^2 (y^2 −1)=4x(x^2 +y^2 ))),((5y^2 (x^2 +1)=3y(x^2 +y^2 ))) :} ⇒5y^2 +5x^2 =(x^2 +y^2 )(3y−4x) ⇒(x^2 +y^2 )(3y−4x−5)=0 x=y=0 or 3y−4x=5 x=((3y−5)/4);(5/4)(3y−5)(y^2 −1)=4y^2 +(1/4)(3y−5)^2 5(3y−5)(y^2 −1)=16y^2 +(3y−5)^2 15y^3 −25y^2 −15y+25=25y^2 +25−30y y(3y^2 −10y^2 +3y)=0 y=0;y=((10−8)/6)=(1/3);y=3 y=3⇒x=1;y=(1/3)⇒x=−1 S={(−1,(1/3));(1,3);(0,0)}

$$\begin{cases}{\mathrm{5}{x}^{\mathrm{2}} \left({y}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{4}{x}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}\\{\mathrm{5}{y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{3}{y}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}\end{cases} \\ $$$$\Rightarrow\mathrm{5}{y}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left(\mathrm{3}{y}−\mathrm{4}{x}\right) \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left(\mathrm{3}{y}−\mathrm{4}{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}={y}=\mathrm{0}\:{or}\:\mathrm{3}{y}−\mathrm{4}{x}=\mathrm{5} \\ $$$${x}=\frac{\mathrm{3}{y}−\mathrm{5}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{3}{y}−\mathrm{5}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{4}{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{y}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}\left(\mathrm{3}{y}−\mathrm{5}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{16}{y}^{\mathrm{2}} +\left(\mathrm{3}{y}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\mathrm{15}{y}^{\mathrm{3}} −\mathrm{25}{y}^{\mathrm{2}} −\mathrm{15}{y}+\mathrm{25}=\mathrm{25}{y}^{\mathrm{2}} +\mathrm{25}−\mathrm{30}{y} \\ $$$${y}\left(\mathrm{3}{y}^{\mathrm{2}} −\mathrm{10}{y}^{\mathrm{2}} +\mathrm{3}{y}\right)=\mathrm{0} \\ $$$${y}=\mathrm{0};{y}=\frac{\mathrm{10}−\mathrm{8}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{3}};{y}=\mathrm{3} \\ $$$${y}=\mathrm{3}\Rightarrow{x}=\mathrm{1};{y}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}=−\mathrm{1} \\ $$$${S}=\left\{\left(−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}}\right);\left(\mathrm{1},\mathrm{3}\right);\left(\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$ \\ $$$$ \\ $$

Commented by Berbere last updated on 13/Mar/24

$$\left(\mathrm{1}\right)\ast{x};\left(\mathrm{2}\right)\ast{y} \\ $$

Commented by cortano12 last updated on 13/Mar/24

oh no. it 5x(y^2 −1) , not 5x^2 (y^2 −1)

$$\mathrm{oh}\:\mathrm{no}.\:\mathrm{it}\:\mathrm{5x}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)\:,\:\mathrm{not}\:\mathrm{5x}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right) \\ $$

Commented by A5T last updated on 13/Mar/24

If x,y∈ C x^2 +y^2 =0⇒5x(y^2 −1)=0 ∧ 5y(x^2 +1)=0 ⇒x=0 or y=+_− 1 ∧ y=0 or x=+_− i (x,y)=(0,0), (i,1),(−i,1),(i,−1),(−i,−1),

$${If}\:{x},{y}\in\:\mathbb{C} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{5}{x}\left({y}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:\wedge\:\mathrm{5}{y}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{or}\:{y}=\underset{−} {+}\mathrm{1}\:\:\:\:\wedge\:{y}=\mathrm{0}\:{or}\:{x}=\underset{−} {+}{i} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right),\:\left({i},\mathrm{1}\right),\left(−{i},\mathrm{1}\right),\left({i},−\mathrm{1}\right),\left(−{i},−\mathrm{1}\right), \\ $$

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