<s>[INST] Systems of equations involving linear and quadratic terms Can we solve for $y$ in this system using algebra?
$$\left\{
\begin{aligned}
x^2 - yz &= 3 \\
y^2 - xz &= 4 \\
z^2 - xy &= 5
\end{aligned}
\right.$$
I’ve tried to evaluate it using elimination and it just gives another equation with unknowns.
First I've tried to multiply the first equation by $y$, second by $z$ and third by $x$. I get $x^2 - y^2z = 3y, y^2z - xz^2 = 4z,$ and $z^2x - x^2y=5x$. Simplifying I get $5x + 4z + 3y = 0$. I've tried it again by multiplying the 1st and 3rd equation by $z, x$ and $y$ respectively. I get $5y + 4x + 3z = 0$. I don't know where to get my third equation.
 [/INST] Multiply each equation by $2$ and add all of them to get
$$(x-y)^2+(y-z)^2+(z-x)^2=24.$$
Let $u=x-y, v=y-z, w=z-x$. Then you have
\begin{align*}
u+v+w&=0\\
u^2+v^2+w^2&=24
\end{align*}
Now in your original system of equations, subtract equ 2 from equ 1, equ 3 from equ 2 and equ 1 from equ 3 to get
\begin{align*}
(x-y)(x+y+z) &=-1\\
(y-z)(x+y+z) &=-1\\
(z-x)(x+y+z) &=2
\end{align*}
Clearly $x+y+z \neq 0$, so from here we can conclude that 
\begin{align*}
x-y&=y-z \implies u=v\\
z-x&=-2(y-z) \implies w=-2v\\
\end{align*}
So now plug this in the $u,v,w$ equation s above to get
$$(v)^2+(v)^2+(-2v)^2=24 \implies v =\pm 2.$$
Hopefully now you can solve the rest.
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