Are the following range of values on ${x}_{3}$ correct given a system of inequalities with 3 variables?

Given the system of inequalities:

$\begin{array}{rl}{x}_{3}& \le \frac{{x}_{1}+{x}_{2}-2}{2}\\ {x}_{3}& \ge {x}_{1}+3{x}_{2}\\ {x}_{3}& \ge 1-{x}_{2}\\ -\frac{11}{3}& \le {x}_{2}\le -3\\ 13& \le {x}_{1}<\mathrm{\infty}\end{array}$

I know that ${x}_{1}\in [lowe{r}_{{x}_{1}},uppe{r}_{{x}_{1}})=[13,\mathrm{\infty})$ and ${x}_{2}\in [lowe{r}_{{x}_{2}},uppe{r}_{{x}_{2}}]=[-11/3,-3]$.

In order to find $lowe{r}_{{x}_{3}},uppe{r}_{{x}_{3}}$, I did the following:

$lowe{r}_{{x}_{3}}=max\{{x}_{1}+3{x}_{2},1-{x}_{2}\}=\{uppe{r}_{{x}_{1}}+3uppe{r}_{{x}_{2}},1-uppe{r}_{{x}_{2}}\}=\{\mathrm{\infty},4\}=\mathrm{\infty}$

$uppe{r}_{{x}_{3}}=\frac{lowe{r}_{{x}_{1}}+lowe{r}_{{x}_{2}}-2}{2}=\frac{13-11/3-2}{2}=\frac{11}{3}$

Is the above calculation correct? In general, is $uppe{r}_{{x}_{n}}=f(lowe{r}_{{x}_{1}},lowe{r}_{{x}_{2}},\dots ,lowe{r}_{{x}_{n-1}})$ and is $lowe{r}_{{x}_{n}}=g(uppe{r}_{{x}_{1}},uppe{r}_{{x}_{2}},\dots ,uppe{r}_{{x}_{n-1}})$, where $f$ and $g$ are some real-valued functions?