Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system : 

$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$

My try but I think not complete : 
The system equivalent : 
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}$$
Then from first equation we have :
$$x+1=\frac{20}{y+1}$$
So third equation give : 
$$z+1=3\frac{y+1}{4}$$
Second equation $\implies $ 
$(y+1)^{2}=16$ then $y=3,-5$ 
This mean $x=\frac{17}{3}$ or $x=-6$ also $z=3$ or $z=-3$ 
I'm correct or no ??
 A: The procedure you followed was correct, including getting the right values of $y$. However, you made mistakes for one value of $x$ (i.e., $\frac{17}{3}$ is not correct) and for both your values of $z$.
However, here's a slightly easier way to deal with the equations instead. You can use the symmetry to multiply your $3$ equations together to get
$$\begin{equation}\begin{aligned}
\left((x+1)(y+1)(z+1)\right)^2 & = 3600 \\
(x+1)(y+1)(z+1) & = \pm 60
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You can then divide this by each of your $3$ equations to get
$$z + 1 = \pm 3 \implies z = -4, 2 \tag{2}\label{eq2A}$$
$$x + 1 = \pm 5 \implies x = -6, 4 \tag{3}\label{eq3A}$$
$$y + 1 = \pm 4 \implies y = -5, 3 \tag{4}\label{eq4A}$$
This leads to the solutions $(-6,-5,-4)$ and $(4,3,2)$.
A: Your method of solution is correct. However, you've made a few trivial mistakes in computing the values of $\ x\ $ and $\ z\ $ when $\ y=3\ $, and the value of $\ z\ $ when $\ y=-5\ $, as John Omielan's answer demonstrates.
A: Alternatively, take logs in
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}. $$
Let $l_x=\ln(|x+1|)$, $l_y=\ln(|y+1|)$, and $l_z=\ln(|z+1|)$ and happily solve
$$\begin{array}{ccc|c}
1 & 1 & 0 & \ln(20)\\
0 & 1 & 1 & \ln(12)\\
1 & 0 & 1 & \ln(15)
\end{array}
$$
in just two steps for $l_x$, $l_y$, and $l_z$.
