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Question Number 196971 by dimentri last updated on 05/Sep/23
solve { ((3x^2 −9y=1)),((3y^2 −9x=0)) :}
$$\:\:{solve}\:\begin{cases}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{9}{y}=\mathrm{1}}\\{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{9}{x}=\mathrm{0}}\end{cases} \\ $$
Answered by Frix last updated on 05/Sep/23
y=((3x^2 −1)/9) x^4 −((2x^2 )/3)−27x+(1/9)=0 No nice solution x≈.00411480∧y≈−.111105 x≈3.07275∧y≈3.03616 x≈−1.53843±2.53397i∧y≈−1.46253∓2.59890i
$${y}=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{9}} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}−\mathrm{27}{x}+\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{nice}\:\mathrm{solution} \\ $$$${x}\approx.\mathrm{00411480}\wedge{y}\approx−.\mathrm{111105} \\ $$$${x}\approx\mathrm{3}.\mathrm{07275}\wedge{y}\approx\mathrm{3}.\mathrm{03616} \\ $$$${x}\approx−\mathrm{1}.\mathrm{53843}\pm\mathrm{2}.\mathrm{53397i}\wedge{y}\approx−\mathrm{1}.\mathrm{46253}\mp\mathrm{2}.\mathrm{59890i} \\ $$
Answered by a.lgnaoui last updated on 05/Sep/23
{ ((x^3 +3yx =(1/3) (1))),((y^3 −3xy = 0 (2))) :} ⇒(1)+(2)⇒ x^3 +y^3 =(1/3) (3) y=^3 (√((1/3)−x^3 )) dans l ′equation donne x=(y^2 /3) alors y=^3 (√((1/3)−(y^6 /(27)))) y^3 =(1/3)−(y^6 /(27)) posons y^3 =Y (Y^2 /(27))+Y−(1/3)=0 ⇒Y^2 +27Y−9=0 { ((Y=((−27−(√(27^2 +36)))/2)=((−27−3(√(85)))/2))),((Y=((−27+(√(27^2 +36)))/2)=((−27+3(√(85)))/2))) :} soit y= { ((−^3 (√((3/2)(9+(√(85)) ))))),((−^3 (√((3/2)(9−(√(85)) ))))) :} x=(y^2 /3) = { ((^3 (√(((9+(√(85)) )^2 )/(12))))),((^3 (√(((9−(√(85)) )^2 )/(12))))) :} S= { (^3 (√(((9+(√(85)) )^2 )/(12))) ;−^3 (√((3/2)(9+(√(85)) ))) ); (^3 (√(((9−(√(85)) )^2 )/(12))) ;−^3 (√((3/2)(9−(√(85)) ))) )}
$$\begin{cases}{\mathrm{x}^{\mathrm{3}} +\mathrm{3yx}\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{y}^{\mathrm{3}} −\mathrm{3xy}\:=\:\mathrm{0}\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\:\:\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\:\:\mathrm{y}=^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{x}^{\mathrm{3}} }\:\:\:\: \\ $$$$\mathrm{dans}\:\:\mathrm{l}\:'\mathrm{equation}\:\:\mathrm{donne}\:\:\mathrm{x}=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\mathrm{alors}\:\:\mathrm{y}=^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{y}^{\mathrm{6}} }{\mathrm{27}}}\:\:\:\:\:\mathrm{y}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{y}^{\mathrm{6}} }{\mathrm{27}} \\ $$$$\mathrm{posons}\:\:\:\mathrm{y}^{\mathrm{3}} =\boldsymbol{\mathrm{Y}} \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{Y}}^{\mathrm{2}} }{\mathrm{27}}+\boldsymbol{\mathrm{Y}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0}\:\:\:\Rightarrow\boldsymbol{\mathrm{Y}}^{\mathrm{2}} +\mathrm{27}\boldsymbol{\mathrm{Y}}−\mathrm{9}=\mathrm{0} \\ $$$$\:\:\:\:\begin{cases}{\boldsymbol{\mathrm{Y}}=\frac{−\mathrm{27}−\sqrt{\mathrm{27}^{\mathrm{2}} +\mathrm{36}}}{\mathrm{2}}=\frac{−\mathrm{27}−\mathrm{3}\sqrt{\mathrm{85}}}{\mathrm{2}}}\\{\boldsymbol{\mathrm{Y}}=\frac{−\mathrm{27}+\sqrt{\mathrm{27}^{\mathrm{2}} +\mathrm{36}}}{\mathrm{2}}=\frac{−\mathrm{27}+\mathrm{3}\sqrt{\mathrm{85}}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{soit}\:\:\:\:\:\mathrm{y}=\begin{cases}{−^{\mathrm{3}} \sqrt{\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{9}+\sqrt{\mathrm{85}}\:\right)}}\\{−^{\mathrm{3}} \sqrt{\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{9}−\sqrt{\mathrm{85}}\:\right)}}\end{cases} \\ $$$$\mathrm{x}=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{3}}\:\:\:\:=\begin{cases}{\:^{\mathrm{3}} \sqrt{\frac{\left(\mathrm{9}+\sqrt{\mathrm{85}}\:\right)^{\mathrm{2}} }{\mathrm{12}}}}\\{\:^{\mathrm{3}} \sqrt{\frac{\left(\mathrm{9}−\sqrt{\mathrm{85}}\:\right)^{\mathrm{2}} }{\mathrm{12}}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{S}}=\:\left\{\:\left(\:^{\mathrm{3}} \sqrt{\frac{\left(\mathrm{9}+\sqrt{\mathrm{85}}\:\right)^{\mathrm{2}} }{\mathrm{12}}}\:;−^{\mathrm{3}} \sqrt{\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{9}+\sqrt{\mathrm{85}}\:\right)}\:\right);\right. \\ $$$$\left.\left(^{\mathrm{3}} \sqrt{\frac{\left(\mathrm{9}−\sqrt{\mathrm{85}}\:\right)^{\mathrm{2}} }{\mathrm{12}}}\:\:;−^{\mathrm{3}} \sqrt{\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{9}−\sqrt{\mathrm{85}}\:\right)}\:\right)\right\} \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 05/Sep/23
{ ((3x^2 −9y=1)),((3y^2 −9x=0)) :} { ((x^2 −3y=(1/3))),((y^2 −3x=0)) :} { ((x^3 −3xy=(x/3) how did you get your equation?)),((y^3 −3xy=0)) :}
$$\begin{cases}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{9}{y}=\mathrm{1}}\\{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{9}{x}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3}{y}=\frac{\mathrm{1}}{\mathrm{3}}}\\{{y}^{\mathrm{2}} −\mathrm{3}{x}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{x}^{\mathrm{3}} −\mathrm{3}{xy}=\frac{{x}}{\mathrm{3}}\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{did}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{equation}}?}\\{{y}^{\mathrm{3}} −\mathrm{3}{xy}=\mathrm{0}}\end{cases} \\ $$
Commented by a.lgnaoui last updated on 05/Sep/23
3y^2 −9x=0 ⇒ x=(y^2 /3)
$$\mathrm{3y}^{\mathrm{2}} −\mathrm{9x}=\mathrm{0}\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{3}} \\ $$
Commented by a.lgnaoui last updated on 05/Sep/23
(1)×x (2)×y ⇒systme.....(1)x+(2)y
$$\left(\mathrm{1}\right)×\mathrm{x} \\ $$$$\left(\mathrm{2}\right)×\mathrm{y}\:\:\:\:\:\:\Rightarrow\mathrm{systme}…..\left(\mathrm{1}\right)\mathrm{x}+\left(\mathrm{2}\right)\mathrm{y} \\ $$
Commented by Frix last updated on 05/Sep/23
But { ((x^3 +3yx =(1/3) (1))),((y^3 −3xy = 0 (2))) :} is wrong
$$\mathrm{But}\:\begin{cases}{\mathrm{x}^{\mathrm{3}} +\mathrm{3yx}\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\mathrm{y}^{\mathrm{3}} −\mathrm{3xy}\:=\:\mathrm{0}\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by a.lgnaoui last updated on 05/Sep/23
exactly thanks for help
$$\mathrm{exactly}\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{help} \\ $$

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