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Question Number 50580 by behi83417@gmail.com last updated on 17/Dec/18

solve for:  x,y,z  .    x^2 +y^2 =6     x^2 +z^2 =3      x(y−z)=2z

solvefor:x,y,z.x2+y2=6x2+z2=3x(yz)=2z

Answered by peter frank last updated on 17/Dec/18

x^2 +y^2 =6.....(i)  x^2 +z^2 =3.....(ii)  x(y−z)=2z....(iii)  i−ii  y^2 −z^2 =3  (y+z)(y−z)=3....(iv)  x(y−z)=2z....(iii)  (y−z)=((2z)/x).....(v)  sub in eqn (iv)  (y+z)((2z)/x)=3....(iv)  (y+z)2z=3x  2zy+2z^2 =3x....(vi)  from  x(y−z)=2z....(iii)  z=((xy)/(x+2))......(vii)  sub in  (ii)  x^2 +z^2 =3.....(ii)  x^2 +((x^2 y^2 )/((x+2)^2 ))=3  .......

x2+y2=6.....(i)x2+z2=3.....(ii)x(yz)=2z....(iii)iiiy2z2=3(y+z)(yz)=3....(iv)x(yz)=2z....(iii)(yz)=2zx.....(v)subineqn(iv)(y+z)2zx=3....(iv)(y+z)2z=3x2zy+2z2=3x....(vi)fromx(yz)=2z....(iii)z=xyx+2......(vii)subin(ii)x2+z2=3.....(ii)x2+x2y2(x+2)2=3.......

Answered by ajfour last updated on 17/Dec/18

let  x=6cos θ, y=6sin θ          x=3cos φ, z=3sin φ  (√6)cos θ((√6)sin θ−(√3)sin φ)=2(√3)sin φ  and    x = (√6)cos θ = (√3)cos φ  ⇒ sin^2 φ = (((6sin θcos θ)/(2(√3)+3(√2)cos θ)))^2 =1−2cos^2 θ  let  cos θ = t  ⇒  36t^2 (1−t^2 )=(2(√3)+3(√2)t)^2 (1−2t^2 )  ⇒  18t^2  = (1−2t^2 )(12+12(√6)t)  ⇒   3t^2 = 6−12t^2 +2(√6)t−4(√6)t^3        4(√6)t^3 +15t^2 −2(√6)t−6 = 0     t ≈ 0.65573  ( the positive root)  x=6t  ,  y=6(√(1−t^2 ))  ,   z = 3(√(1−2t^2 )) .

letx=6cosθ,y=6sinθx=3cosϕ,z=3sinϕ6cosθ(6sinθ3sinϕ)=23sinϕandx=6cosθ=3cosϕsin2ϕ=(6sinθcosθ23+32cosθ)2=12cos2θletcosθ=t36t2(1t2)=(23+32t)2(12t2)18t2=(12t2)(12+126t)3t2=612t2+26t46t346t3+15t226t6=0t0.65573(thepositiveroot)x=6t,y=61t2,z=312t2.

Commented by behi83417@gmail.com last updated on 17/Dec/18

(2(√3)+3(√2)t)^2    ?

(23+32t)2?

Commented by peter frank last updated on 17/Dec/18

please sir Ajfour  insert some steps

pleasesirAjfourinsertsomesteps

Commented by ajfour last updated on 17/Dec/18

did..

did..

Commented by peter frank last updated on 17/Dec/18

thank you

thankyou

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