# 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.

• What do you think? – manooooh Jun 18 at 5:22
• I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Jun 18 at 5:22
• Hello, Ive tried to evaluate it using ellimination and it just give another equations with unknowns. – Jan Navasca Jun 18 at 6:02
• @Jan Navasca Show it please. Show at least two first steps. – Michael Rozenberg Jun 18 at 6:10
• First Ive tried to multiply the first equation by y 2nd 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. Ive tried it again by multiplying the 1st wnd and 3rd equation by z, x and y respectively. I get 5y + 4x + 3z = 0. I dont know where to get my third equation – Jan Navasca Jun 18 at 6:14

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.

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

• We did the same for the 1st and 2nd equation. What do you mean the rest is smooth? How'd you get it? – Jan Navasca Jun 18 at 6:22
• @Jan Navasca I added something. See now. – Michael Rozenberg Jun 18 at 6:26
• Thank you so much! You're amazing! – Jan Navasca Jun 18 at 6:35
• @Jan Navasca You are welcome! – Michael Rozenberg Jun 18 at 6:37

$$$$x^2-yz=3\hspace{2cm}(1)\\ y^2-xz=4\hspace{2cm}(2)\\ z^2-xy=5\hspace{2cm}(3)$$$$ $$(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 $$x=\frac{-11}{6}, y=\frac{1}{6}, z=\frac{13}{6}$$

• You don't need to to check the case $x+y+z=0$ as it will violate your equation (4). – Anurag A Jun 18 at 7:02
• @AnuragA Thank you so much. I failed to observe that. But that method I used may help someone in some other problem. – Always a Learner Jun 18 at 7:09