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Question Number 135191 by liberty last updated on 11/Mar/21

{ ((x^2 −yz = 3)),((y^2 − zx = 5)),((z^2 −xy = −1)) :} solve for x ,y and z.

{x2−yz=3y2−zx=5z2−xy=−1solveforx,yandz.

Answered by MJS_new last updated on 11/Mar/21

y=px∧z=qx (1) (1−pq)x^2 −3=0 ⇒ x^2 =(3/(1−pq)) (2) (p^2 −q)x^2 −5=0 ⇒ x^2 =(5/(p^2 −q)) (3) (q^2 −p)x^2 +1=0 ⇒ x^2 =(1/(p−q^2 )) (a) (3/(1−pq))=(5/(p^2 −q)) (b) (3/(1−pq))=(1/(p−q^2 )) (a) 3p^2 +5pq−3q−5=0 ⇒ q=((5−3p^2 )/(5p−3)) (b) pq+3p−3q^2 −1=0 inserting into (b) leads to p^4 −2p^3 −p+2=0 (p−2)(p−1)(p^2 +p+1)=0 p=1∨p=2∨p=−(1/2)±((√3)/2)i q=1∨q=−1∨q=−(1/2)∓((√3)/2)i the only possible solution is x^2 =1∧y=2x∧z=−x ⇒ x=−1∧y=−2∧z=1 ∨ x=1∧y=2∧z=−1

y=px∧z=qx(1)(1−pq)x2−3=0⇒x2=31−pq(2)(p2−q)x2−5=0⇒x2=5p2−q(3)(q2−p)x2+1=0⇒x2=1p−q2(a)31−pq=5p2−q(b)31−pq=1p−q2(a)3p2+5pq−3q−5=0⇒q=5−3p25p−3(b)pq+3p−3q2−1=0insertinginto(b)leadstop4−2p3−p+2=0(p−2)(p−1)(p2+p+1)=0p=1∨p=2∨p=−12±32iq=1∨q=−1∨q=−12∓32itheonlypossiblesolutionisx2=1∧y=2x∧z=−x⇒x=−1∧y=−2∧z=1∨x=1∧y=2∧z=−1

Answered by benjo_mathlover last updated on 11/Mar/21

adding the three equation gives ⇒x^2 +y^2 +z^2 −(xy+xz+yz)=7 ⇒2(x^2 +y^2 +z^2 )−2(xy+xz+yz)=14 let u=x−y , v=y−z and w=z−x we get u^2 +v^2 +w^2 = 14 while obviously u+v+w = 0 w = −(u+v) substituing gives u^2 +v^2 +uv = 7 .now substract the equation y^2 −zx=5 from first one in the original problem x^2 −y^2 +zx−zy = −2 (x−y)(x+y+z)=−2...(i) ⇒the third from the second gives (y−z)(x+y+z)=6 ...(2) (u,v)=(x−y,y−z) we see that v=−3u and u^2 =1 so (u,v)=(1,−3)or (−1,3) . case(1) { ((x=y+1)),((z=y+3)) :} substituting the equation y^2 =5+zx we find the solution { (((−1,−2,1) and)),(((1,2,−1))) :}

addingthethreeequationgives⇒x2+y2+z2−(xy+xz+yz)=7⇒2(x2+y2+z2)−2(xy+xz+yz)=14letu=x−y,v=y−zandw=z−xwegetu2+v2+w2=14whileobviouslyu+v+w=0w=−(u+v)substituinggivesu2+v2+uv=7.nowsubstracttheequationy2−zx=5fromfirstoneintheoriginalproblemx2−y2+zx−zy=−2(x−y)(x+y+z)=−2...(i)⇒thethirdfromthesecondgives(y−z)(x+y+z)=6...(2)(u,v)=(x−y,y−z)weseethatv=−3uandu2=1so(u,v)=(1,−3)or(−1,3).case(1){x=y+1z=y+3substitutingtheequationy2=5+zxwefindthesolution{(−1,−2,1)and(1,2,−1)

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