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

$$\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{f}\mathit{o}\mathit{r}:\mathit{x},\mathit{y},\mathit{z}.$$$${\mathbf{x}}^2+{\mathbf{y}}^2=6$$$${\mathbf{x}}^2+{\mathbf{z}}^2=3$$$$\mathbf{x}(\mathbf{y}-\mathbf{z})=2\mathbf{z}$$

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

$$x^2+y^2=6.....\left(i\right)$$$$x^2+z^2=3.....\left(ii\right)$$$$x(y-z)=2z....\left(iii\right)$$$$i-ii$$$$y^2-z^2=3$$$$(y+z)(y-z)=3....\left(iv\right)$$$$x(y-z)=2z....\left(iii\right)$$$$(y-z)=\frac{2z}{x}.....\left(v\right)$$$$subineqn\left(iv\right)$$$$(y+z)\frac{2z}{x}=3....\left(iv\right)$$$$(y+z)2z=3x$$$$2zy+2z^2=3x....\left(vi\right)$$$$from$$$$x(y-z)=2z....\left(iii\right)$$$$z=\frac{xy}{x+2}......\left(vii\right)$$$$subin\left(ii\right)$$$$x^2+z^2=3.....\left(ii\right)$$$$x^2+\frac{x^2{y}^2}{{(x+2)}^2}=3$$$$.......$$$$$$

Answered by ajfour last updated on 17/Dec/18

$$letx=6\cos \theta ,y=6\sin \theta$$$$x=3\cos \varphi ,z=3\sin \varphi$$$$\sqrt{6}\cos \theta (\sqrt{6}\sin \theta -\sqrt{3}\sin \varphi )=2\sqrt{3}\sin \varphi$$$$andx=\sqrt{6}\cos \theta =\sqrt{3}\cos \varphi$$$$\Rightarrow \sin {}^2\varphi ={\left(\frac{6\sin \theta \cos \theta }{2\sqrt{3}+3\sqrt{2}\cos \theta }\right)}^2=1-2\cos {}^2\theta$$$$let\cos \theta =t$$$$\Rightarrow 36t^2(1-t^2)={(2\sqrt{3}+3\sqrt{2}t)}^2(1-2t^2)$$$$\Rightarrow 18t^2=(1-2t^2)(12+12\sqrt{6}t)$$$$\Rightarrow 3t^2=6-12t^2+2\sqrt{6}t-4\sqrt{6}{t}^3$$$$4\sqrt{6}{t}^3+15t^2-2\sqrt{6}t-6=0$$$$t\approx 0.65573\left(thepositiveroot\right)$$$$x=6t,y=6\sqrt{1-t^2},$$$$z=3\sqrt{1-2t^2}.$$

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

$${(2\sqrt{3}+3\sqrt{2}t)}^2?$$

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

$$pleasesirAjfourinsertsomesteps$$

Commented by ajfour last updated on 17/Dec/18

$$did..$$

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

$$thankyou$$

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