System of equations with multiplication

Question

I want to find all $x,y,z\in\mathbb{R}$ such that $(x+1)yz=12, (y+1)zx=4, (z+1)xy=4$.

I can multiply all three equations to get $(x+1)(y+1)(z+1)x^2y^2z^2=192$.

I can divide the first equation by the second to get $\dfrac{(x+1)y}{(y+1)x}=4$, which simplifies to $3xy+4x-y=0$.

None of these seems to help. How can I solve the equations?

Answer 1

Your equations are

$$\begin{align*}I&\;\;xyz+yz&=12\\ II&\;\;xyz+xz&=4\\ III&\;\;xyz+xy&=4\end{align*}$$

Well, now a little algebra:

$$\begin{align*}I-II:&\;\;(x-y)z=8\\ I-III:&\;\;(x-z)y=8\\ II-III:&\;\;(z-y)x=0\end{align*}$$

Since $\;x\neq 0\;$ (why?), we get $\;y=z\;$ and thus $\;xz=xy=y^2+8\;$ ...take it from here.

Answer 2

HINT:

Clearly, $xyz\ne0$

From the last two relation we have $4-xyz=zx=xy\implies y=z$

The division of first equation by the second gives $$\frac{(x+1)y}{x(y+1)}=\frac{12}4$$

Please check the cases of division by $0$