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 May 9th, 2018, 03:20 AM #1 Newbie   Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 System of equations \begin{aligned} x^2+xy+y^2=1\\ x^2+xz+z^2=4\\ y^2+zy+z^2=7 \end{aligned} Subtracting 1st from 2nd, and 2nd from 3rd: \begin{aligned} &(z-y)(x+y+z)=3\\ &(y-x)(x+y+z)=3\\ &\implies z-y=y-x\Leftrightarrow z=2y-x \end{aligned} How do I proceed from here? I've been at it for quite a while, and I still feel like I'm just short of untangling it. Thanks.
 May 9th, 2018, 03:44 AM #2 Member   Joined: Jan 2018 From: Belgrade Posts: 54 Thanks: 2 Try to do this: $\displaystyle (y-x) \cdot 3y=3$, that is: $\displaystyle (y-x) \cdot y=1$. From here: $\displaystyle x=y-\frac{1}{y}$. Put it in the first equation, and you get: $\displaystyle 3y^{2}+\frac{1}{y^{2}}=4$ Substitute $\displaystyle t=y^{2}$ and you have: $\displaystyle 3t^{2}-4t+1=0$ Thanks from bongcloud

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