Resoudre le systeme d′ equations: { (((x+y)xy=84)),((x^2 +y^2 =25)) :}


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