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Question Number 17985 by tawa tawa last updated on 13/Jul/17
solve simultaneously.  x^3  + y^3  = 35  x^4  + y^4  = 97
solvesimultaneously.x3+y3=35x4+y4=97
Answered by mrW1 last updated on 13/Jul/17
let u=x+y and xy=v  (x+y)(x^2 +y^2 +2xy−3xy)=35  (x+y)[(x+y)^2 −3xy]=35  u(u^2 −3v)=35  v=((u^2 −((35)/u))/3)=((u^3 −35)/(3u))    (x^2 )^2 +(y^2 )^2 +2x^2 y^2 −2x^2 y^2 =97  (x^2 +y^2 )^2 −2x^2 y^2 =97  (x^2 +y^2 +2xy−2xy)^2 −2x^2 y^2 =97  ((x+y)^2 −2xy)^2 −2x^2 y^2 =97  (u^2 −2v)^2 −2v^2 =97  (u^2 −2((u^3 −35)/(3u)))^2 −2(((u^3 −35)/(3u)))^2 =97  (u^3 +70)^2 −2(u^3 −35)^2 =873u^2   u^6 +140u^3 +4900−2u^6 +140u^3 −2450=873u^2   u^6 −280u^3 +873u^2 −2450=0  ⇒u=−1.391 or 5  ⇒v=9.0322 or 6    x+y=u  xy=v  x^2 +y^2 −2xy+4xy=u^2   (x−y)^2 =u^2 −4v  x−y=±(√(u^2 −4v))  ⇒x=((u±(√(u^2 −4v)))/2)  ⇒y=((u∓(√(u^2 −4v)))/2)    with u=−1.391 and v=9.0322  there is no real solution since  u^2 −4v<0    with u=5 and v=6  ⇒x+y=5  ⇒x−y=±(√(25−24))=±1  ⇒x=((5±1)/2)=3,2  ⇒y=((5∓1)/2)=2,3    ⇒(x,y)=(3,2) or (2,3)
letu=x+yandxy=v(x+y)(x2+y2+2xy3xy)=35(x+y)[(x+y)23xy]=35u(u23v)=35v=u235u3=u3353u(x2)2+(y2)2+2x2y22x2y2=97(x2+y2)22x2y2=97(x2+y2+2xy2xy)22x2y2=97((x+y)22xy)22x2y2=97(u22v)22v2=97(u22u3353u)22(u3353u)2=97(u3+70)22(u335)2=873u2u6+140u3+49002u6+140u32450=873u2u6280u3+873u22450=0u=1.391or5v=9.0322or6x+y=uxy=vx2+y22xy+4xy=u2(xy)2=u24vxy=±u24vx=u±u24v2y=uu24v2withu=1.391andv=9.0322thereisnorealsolutionsinceu24v<0withu=5andv=6x+y=5xy=±2524=±1x=5±12=3,2y=512=2,3(x,y)=(3,2)or(2,3)
Commented by tawa tawa last updated on 13/Jul/17
God bless you sir. i really appreciate.
Godblessyousir.ireallyappreciate.

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