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.
 A: 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.
A: 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:
$$\left\{\left(\frac{11}{6},-\frac{1}{6},-\frac{13}{6}\right),\left(-\frac{11}{6},\frac{1}{6},\frac{13}{6}\right)\right\}.$$
A: \begin{equation}
x^2-yz=3\hspace{2cm}(1)\\
y^2-xz=4\hspace{2cm}(2)\\
z^2-xy=5\hspace{2cm}(3)
\end{equation}
$(2)-(1)\implies$
$$(y-x)\cdot(x+y+z)=1\hspace{2cm}(4)$$
$(3)-(2)\implies$
$$(z-y)\cdot(x+y+z)=1\hspace{2cm}(5)$$
$(5)-(4)\implies$
$$(x+y+z)\cdot(2y-x-z)=0\hspace{2cm}(6)$$
$\implies$
$$x=-(y+z)$$ or $$x=(2y-z)$$
Now suppose $(x+y+z)=0$,Then
$(1)\implies$ $$(y+z)^2-yz=3
\implies y^2+z^2+yz=3$$
$(2)+(3)\implies$
$$y^2+z^2+(y+z)(y+z)=9$$
$\implies$
$$2(y^2+z^2+yz)=9 \implies 2\cdot 3=9 \hspace{2cm}\Rightarrow\Leftarrow
$$
So $x=2y-z$
$(2)\implies$
\begin{align}
y^2-(2y-z)z=4 \implies (y-z)^2=4\implies y=z\pm 2\hspace{2cm}(7)\\
\end{align}
Also since $z=2y-x$
\begin{align}
y^2-(2y-x)x=4 \implies (x-y)^2=4 \implies x=y\pm 2\hspace{2cm}(8)\\
\end{align}
If $x=z$, then
$(1)\implies$ $$x^2-xy=3$$
$(3)\implies$ $$x^2-xy=5 \hspace{2cm} \Rightarrow\Leftarrow $$
Therefore $x\neq z$ and the possible combinations are $(x,x+2,x+4)$ and $(x,x-2,x-4)$[from $(7)\&(8)$]
Assuming $y=z+2$ and solving $(3)$, we get $$x=\frac{11}{6}, y=\frac{-1}{6}, z=\frac{-13}{6}$$
Assuming $y=z-2$ and solving $(3)$, we get $$x=\frac{-11}{6}, y=\frac{1}{6}, z=\frac{13}{6}$$
