<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] We have
$$\left\{
\begin{aligned}
x^2y - y^2z &= 3y \\
y^2z - xz^2 &= 4z \\
z^2x - x^2y &= 5x
\end{aligned}
\right.
$$ and after summing we obtain:
$$3y+4z+5x=0.$$
Also,
$$
\left\{
\begin{aligned}
x^2z - yz^2 &= 3z \\
y^2x - x^2z &= 4x \\
z^2y - xy^2 &= 5y
\end{aligned}
\right.
$$
and after summing again we obtain:
$$3z+4x+5y=0.$$
The rest is smooth:
From $$5x+3y+4z=0$$ and
$$4x+5y+3z=0$$ we obtain:
$$y=-\frac{x}{11}$$ and $$z=-\frac{13x}{11},$$ which gives
$$x^2-\frac{13x^2}{121}=3$$ and from here $$x=\pm\frac{11}{6}.$$
I got the following answer:
$$\left\{\left(\frac{11}{6},-\frac{1}{6},-\frac{13}{6}\right),\left(-\frac{11}{6},\frac{1}{6},\frac{13}{6}\right)\right\}.$$
</s>