How does one solve the following system of equations?

Question

$\left\{ \begin{aligned} |x| + |y| + z &= a\\ |x| + y + |z| &= a + 1\\ x + |y| + |z| &= a + 2\end{aligned} \right.$ The domain is $\mathbb{R}$, $a > 1$.

Answer 1

Hint...consider separately the cases when each of $x$, $y$ and $z$ are $\geq 0$ or $<0$. There are eight cases to consider. In all but two cases, the equations are inconsistent.

the solutions are $$(a+\frac 12,-\frac 12, -1)$$ and $$(-a-\frac 12,-a-1,-a-\frac 32)$$

Answer 2

From the first and from the third equations we obtain: $$x+|y|+|z|=|x|+|y|+z+2,$$ which gives $$|z|-z=|x|-x+2\geq2$$ and from here $$|z|\geq2+z$$ and since $$z\geq z+2$$ is impossible, we obtain: $$z\leq -z-2$$ or $$z\leq-1.$$ Now, from the first and from the second equations we obtain: $$|x|+y+|z|=|x|+|y|+z+1,$$ which gives $$|y|-y=|z|-z-1\geq1+1-1\geq1,$$ which gives $$|y|\geq1+y,$$ $$y\leq-1-y$$ or $$y\leq-\frac{1}{2}.$$ Id est, we got the following system: $$|x|-y+z=a,$$ $$|x|+y-z=a+1$$ and $$x-y-z=a+2,$$ which gives $$|x|=a+\frac{1}{2}.$$ Can you end it now?

I got the following answer. $$\left\{\left(a+\frac{1}{2},-\frac{1}{2},-1\right)\right\}$$