{ ((xy+x+y = 20)),((x^2 +y^2 = 40)) :}find x and y


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Question Number 105158 by bemath last updated on 26/Jul/20
${ ((xy+x+y = 20)),((x^2 +y^2 = 40)) :}find x and y$
${\begin{cases} x y + x + y = 20 \\ x^{2} + y^{2} = 40 \end{cases} f i n d x a n d y$
	Answered by Dwaipayan Shikari last updated on 26/Jul/20
	$xy=20−(x+y) x^2 +y^2 +2xy=(x+y)^2 40+40−2(x+y)=(x+y)^2 (x+y)^2 +2(x+y)−80=0 x+y=((−2±(√(4+320)))/2)=8 or −10 (x+y)^2 −2xy=x^2 +y^2 64−2xy=40 xy=12 (x+y)^2 −4xy=(x−y)^2 16=(x−y)^2 x−y=±4 and x+y=8 or −10 { ((x=6,2)),((y=2,6)) :}$
	$x y = 20 - (x + y)$ $x^{2} + y^{2} + 2 x y = {(x + y)}^{2}$ $40 + 40 - 2 (x + y) = {(x + y)}^{2}$ ${(x + y)}^{2} + 2 (x + y) - 80 = 0$ $x + y = \frac{- 2 \pm \sqrt{4 + 320}}{2} = 8 o r - 10$ ${(x + y)}^{2} - 2 x y = x^{2} + y^{2}$ $64 - 2 x y = 40$ $x y = 12$ ${(x + y)}^{2} - 4 x y = {(x - y)}^{2}$ $16 = {(x - y)}^{2}$ $x - y = \pm 4 a n d x + y = 8 o r - 10$ ${\begin{cases} x = 6, 2 \\ y = 2, 6 \end{cases}$
	Answered by bramlex last updated on 26/Jul/20
	$⇒(x+y)^2 −2xy = 40 ⇒x+y = 20−xy ⇒(20−xy)^2 −2xy = 40 let xy = m ⇒ (20−m)^2 −2m =40 400−40m+m^2 −2m−40=0 m^2 −42m+360 = 0 m = ((42 ± (√(324)))/2) = ((42±18)/2) m = 21±9 → { ((m_1 = 30)),((m_2 = 12)) :} for m_1 = xy = 30→x+y = −10 ⇒x(−10−x)= 30 ; x^2 +10x+30=0 x_(1,2) = ((−10±2i(√5))/2) = −5 ± i(√5) y_(1,2) = −10+5∓i(√5) = −5∓i(√5) for m_2 = xy = 12→x+y = 8 ⇒x(8−x) = 12 ; x^2 −8x+12 = 0 x_(3,4) = ((8±4)/2) = 4±2→ { ((x_3 = 6)),((x_4 = 2)) :} y_(3,4) = 4∓2 → { ((y_3 = 2)),((y_4 = 6)) :}$
	$\Rightarrow {(x + y)}^{2} - 2 x y = 40$ $\Rightarrow x + y = 20 - x y$ $\Rightarrow {(20 - x y)}^{2} - 2 x y = 40$ $l e t x y = m \Rightarrow {(20 - m)}^{2} - 2 m = 40$ $400 - 40 m + m^{2} - 2 m - 40 = 0$ $m^{2} - 42 m + 360 = 0$ $m = \frac{42 \pm \sqrt{324}}{2} = \frac{42 \pm 18}{2}$ $m = 21 \pm 9 \to {\begin{cases} m_{1} = 30 \\ m_{2} = 12 \end{cases}$ $f o r m_{1} = x y = 30 \to x + y = - 10$ $\Rightarrow x (- 10 - x) = 30; x^{2} + 10 x + 30 = 0$ $x_{1, 2} = \frac{- 10 \pm 2 i \sqrt{5}}{2} = - 5 \pm i \sqrt{5}$ $y_{1, 2} = - 10 + 5 \mp i \sqrt{5} = - 5 \mp i \sqrt{5}$ $f o r m_{2} = x y = 12 \to x + y = 8$ $\Rightarrow x (8 - x) = 12; x^{2} - 8 x + 12 = 0$ $x_{3, 4} = \frac{8 \pm 4}{2} = 4 \pm 2 \to {\begin{cases} x_{3} = 6 \\ x_{4} = 2 \end{cases}$ $y_{3, 4} = 4 \mp 2 \to {\begin{cases} y_{3} = 2 \\ y_{4} = 6 \end{cases}$
	Answered by behi83417@gmail.com last updated on 26/Jul/20
	$x+y=p,xy=q ⇒ { ((p+q=20)),((p^2 −2q=40⇒p^2 −2(20−p)=40)) :} ⇒p^2 +2p+1=81⇒p+1=±9 { ((p=8,−10)),((q=12,30)) :} ⇒z^2 −8z+12=0⇒z=2,6⇒(x,y)=(2,6) ⇒z^2 +10z+30=0⇒z=((−10±(√(100−120)))/2) ⇒z=−5±i(√5)⇒(x,y)=(−5+i(√5),−5−i(√5)) ■$
	$x + y = p, xy = q$ $\Rightarrow {\begin{cases} p + q = 20 \\ p^{2} - 2 q = 40 \Rightarrow p^{2} - 2 (20 - p) = 40 \end{cases}$ $\Rightarrow p^{2} + 2 p + 1 = 81 \Rightarrow p + 1 = \pm 9$ ${\begin{cases} p = 8, - 10 \\ q = 12, 30 \end{cases}$ $\Rightarrow z^{2} - 8 z + 12 = 0 \Rightarrow z = 2, 6 \Rightarrow (x, y) = (2, 6)$ $\Rightarrow z^{2} + 10 z + 30 = 0 \Rightarrow z = \frac{- 10 \pm \sqrt{100 - 120}}{2}$ $\Rightarrow z = - 5 \pm i \sqrt{5} \Rightarrow (x, y) = (- 5 + i \sqrt{5}, - 5 - i \sqrt{5}) ◼$
	Answered by mathmax by abdo last updated on 26/Jul/20
	$let x+y =u and xy =v ⇒ { ((u+v =20)),((u^2 −2v =40 ⇒)) :} { ((u =20−v)),(((20−v)^2 −2v =40)) :} (20−v)^2 −2v =40 ⇒400−40v +v^2 −2v−40 =0 ⇒ v^2 −42v+360 =0 Δ^′ =21^2 −360 =441−360 =81 ⇒v_1 =21+9 =20 and v_2 =21−9 =12 we have u =20−v ⇒u =0 or u =8 for v =20 we get u =0 ⇒ { ((x+y =0)),((xy =20)) :} ⇒ { ((y=−x)),((−x^2 =20 (impossible))) :} for v =12 ⇒u =8 ⇒ { ((x+y =8)),((xy =12)) :} x and y are solution of t^2 −8t +12 =0 Δ^′ =4^2 −12 =4 ⇒t_1 =4+2 =6 and t_2 =4−2 =2 ⇒ (x,y) =(6,2) or (x,y) =(2,6)(system is symetric)$
	$let x + y = u and xy = v \Rightarrow {\begin{cases} u + v = 20 \\ u^{2} - 2 v = 40 \Rightarrow \end{cases}$ ${\begin{cases} u = 20 - v \\ {(20 - v)}^{2} - 2 v = 40 \end{cases}$ ${(20 - v)}^{2} - 2 v = 40 \Rightarrow 400 - 40 v + v^{2} - 2 v - 40 = 0 \Rightarrow$ $v^{2} - 42 v + 360 = 0$ $Δ^{^{'}} = 21^{2} - 360 = 441 - 360 = 81$ $\Rightarrow v_{1} = 21 + 9 = 20 and v_{2} = 21 - 9 = 12$ $we have u = 20 - v \Rightarrow u = 0 or u = 8$ $for v = 20 we get u = 0 \Rightarrow {\begin{cases} x + y = 0 \\ xy = 20 \end{cases}$ $\Rightarrow {\begin{cases} y = - x \\ - x^{2} = 20 (impossible) \end{cases}$ $for v = 12 \Rightarrow u = 8 \Rightarrow {\begin{cases} x + y = 8 \\ xy = 12 \end{cases}$ $x and y are solution of t^{2} - 8 t + 12 = 0$ $Δ^{^{'}} = 4^{2} - 12 = 4 \Rightarrow t_{1} = 4 + 2 = 6 and t_{2} = 4 - 2 = 2 \Rightarrow$ $(x, y) = (6, 2) or (x, y) = (2, 6) (system is symetric)$
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