Systems of equations involving linear and quadratic terms

Question

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.

Answer 1

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.

Answer 2

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: $$\left\{\left(\frac{11}{6},-\frac{1}{6},-\frac{13}{6}\right),\left(-\frac{11}{6},\frac{1}{6},\frac{13}{6}\right)\right\}.$$

Answer 3

\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 $$x=\frac{-11}{6}, y=\frac{1}{6}, z=\frac{13}{6}$$