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 ??

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.
