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Question Number 213721 by a.lgnaoui last updated on 14/Nov/24
Resoudre le systeme d′ equations: { (((x+y)xy=84)),((x^2 +y^2 =25)) :}
$$\boldsymbol{\mathrm{Resoudre}}\:\boldsymbol{\mathrm{le}}\:\boldsymbol{\mathrm{systeme}}\:\boldsymbol{\mathrm{d}}'\:\boldsymbol{\mathrm{equations}}: \\ $$$$\begin{cases}{\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{xy}}=\mathrm{84}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:\:\:\:\:\:=\mathrm{25}}\end{cases} \\ $$
Answered by golsendro last updated on 14/Nov/24
(x+y)^2 −2xy= x^2 +y^2 =25 Let { ((x+y=a)),((xy=b)) :} { ((ab=84)),((a^2 −2b=25⇒a^3 −2ab=25a)) :} a^3 −25a−168=0 (a−7)(a^2 +7a+24)=0 a=7 ⇒b=12 { ((x+y=7)),((xy=12)) :}
$$\:\:\:\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{2xy}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{25} \\ $$$$\:\:\:\mathrm{Let}\:\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{a}}\\{\mathrm{xy}=\mathrm{b}}\end{cases} \\ $$$$\:\:\:\begin{cases}{\mathrm{ab}=\mathrm{84}}\\{\mathrm{a}^{\mathrm{2}} −\mathrm{2b}=\mathrm{25}\Rightarrow\mathrm{a}^{\mathrm{3}} −\mathrm{2ab}=\mathrm{25a}}\end{cases} \\ $$$$\:\:\:\:\:\mathrm{a}^{\mathrm{3}} −\mathrm{25a}−\mathrm{168}=\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{a}−\mathrm{7}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{7a}+\mathrm{24}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{a}=\mathrm{7}\:\Rightarrow\mathrm{b}=\mathrm{12} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{7}}\\{\mathrm{xy}=\mathrm{12}}\end{cases} \\ $$
Answered by Frix last updated on 14/Nov/24
x=u−(√v)∧y=u+(√v) { ((2u(u^2 −v)=84)),((2(u^2 +v)=25)) :} { ((v=((u^3 −42)/u))),((v=((25−2u^2 )/2))) :} ((u^3 −42)/u)=((25−2u^2 )/2) u^3 −((25)/4)u−21=0 u=(7/2)∨u=−(7/4)±((√(47))/4)i ⇒ v=(1/4)∨v=((99)/8)±((7(√(47)))/8)i ⇒ x=3∧y=4 It′s possible to give the exact forms of the complex solutions but they′re not “nice” x≈−5.36437±.884073 y≈1.86437±2.54375 And of course we can exchange x⇄y
$${x}={u}−\sqrt{{v}}\wedge{y}={u}+\sqrt{{v}} \\ $$$$\begin{cases}{\mathrm{2}{u}\left({u}^{\mathrm{2}} −{v}\right)=\mathrm{84}}\\{\mathrm{2}\left({u}^{\mathrm{2}} +{v}\right)=\mathrm{25}}\end{cases} \\ $$$$\begin{cases}{{v}=\frac{{u}^{\mathrm{3}} −\mathrm{42}}{{u}}}\\{{v}=\frac{\mathrm{25}−\mathrm{2}{u}^{\mathrm{2}} }{\mathrm{2}}}\end{cases} \\ $$$$\frac{{u}^{\mathrm{3}} −\mathrm{42}}{{u}}=\frac{\mathrm{25}−\mathrm{2}{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{25}}{\mathrm{4}}{u}−\mathrm{21}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{7}}{\mathrm{2}}\vee{u}=−\frac{\mathrm{7}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{47}}}{\mathrm{4}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{4}}\vee{v}=\frac{\mathrm{99}}{\mathrm{8}}\pm\frac{\mathrm{7}\sqrt{\mathrm{47}}}{\mathrm{8}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{3}\wedge{y}=\mathrm{4} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{give}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{forms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{complex}\:\mathrm{solutions}\:\mathrm{but}\:\mathrm{they}'\mathrm{re}\:\mathrm{not}\:“\mathrm{nice}'' \\ $$$${x}\approx−\mathrm{5}.\mathrm{36437}\pm.\mathrm{884073} \\ $$$${y}\approx\mathrm{1}.\mathrm{86437}\pm\mathrm{2}.\mathrm{54375} \\ $$$$\mathrm{And}\:\mathrm{of}\:\mathrm{course}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange}\:{x}\rightleftarrows{y} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Nov/24
{ (((x+y)xy=84....i)),((x^2 +y^2 =25...ii)) :} i⇒ (x+y)^2 (xy)^2 =84^2 ⇒(x+y)^2 =((84^2 )/((xy)^2 ))...iii ii⇒(x+y)^2 =25+2xy...iv iii & iv⇒((84^2 )/((xy)^2 ))=25+2xy ⇒2(xy)^3 +25(xy)^2 −84^2 =0 let xy=z 2z^3 +25z^2 −84^2 =0 (z−12)(2z^2 +49z+588)=0 z=12 or z=((−49±7i(√(47)))/4)
$$\begin{cases}{\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{xy}}=\mathrm{84}….{i}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:\:\:\:\:\:=\mathrm{25}…{ii}}\end{cases} \\ $$$${i}\Rightarrow\:\:\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\mathrm{2}} \left(\boldsymbol{\mathrm{xy}}\right)^{\mathrm{2}} =\mathrm{84}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} =\frac{\mathrm{84}^{\mathrm{2}} }{\left(\mathrm{xy}\right)^{\mathrm{2}} }…{iii} \\ $$$${ii}\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} =\mathrm{25}+\mathrm{2xy}…{iv} \\ $$$${iii}\:\&\:{iv}\Rightarrow\frac{\mathrm{84}^{\mathrm{2}} }{\left(\mathrm{xy}\right)^{\mathrm{2}} }=\mathrm{25}+\mathrm{2xy} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}\left(\mathrm{xy}\right)^{\mathrm{3}} +\mathrm{25}\left(\mathrm{xy}\right)^{\mathrm{2}} −\mathrm{84}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\mathrm{xy}=\mathrm{z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2z}^{\mathrm{3}} +\mathrm{25z}^{\mathrm{2}} −\mathrm{84}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{z}−\mathrm{12}\right)\left(\mathrm{2z}^{\mathrm{2}} +\mathrm{49z}+\mathrm{588}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{z}=\mathrm{12}\:\mathrm{or}\:\mathrm{z}=\frac{−\mathrm{49}\pm\mathrm{7}{i}\sqrt{\mathrm{47}}}{\mathrm{4}} \\ $$

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