Question Number 105158 by bemath last updated on 26/Jul/20

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

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)) :}

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)) ■

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)