<s>[INST] Solve the following system of equations:$|x|+|y|=2$ and $y=x+1$. 
Solve the following system of equations:
$|x|+|y|=2$ and $y=x+1$, where $x$ is a real number.

Approach:
I substituted $y$ in equation $1$, so:
$$ \ \ \ \ \ \ \ \ \ \ |x|+|x+1|=2$$
$$1 \ \ \ \ \ \ \ \ \ \ 1$$
$$2 \ \ \ \ \ \ \ \ \ \ 0$$
$$0 \ \ \ \ \ \ \ \ \ \ 2$$
These are the total possibilities I think, because $|x|, |x+1| \ge 0$.
Then I made cases,
Case $1$: $|x|=1$, so, $x = \pm 1$
and $|x+1| = 1$, so $x+1=\pm 1$ so, $x = 0,-2$.
But no value of $x$ is matching, so this case gets rejected.
Case $2$: $|x|=2$, so, $x = \pm 2$
and $|x+1| = 0$, so $x=-1$.
But no value of $x$ is matching, so this case also gets rejected.
Case $3$: $|x|=0$, so, $x = 0$
and $|x+1| = 2$, so $x+1 = \pm 2$ and so, $x=1,-3$.
Here also, no value of $x$ is matching. So no solution exists.
Is this solution correct? Please confirm. If there is a shorter method to approach the question, please share it.
 [/INST] HINT

*

*If $x≥0$, then $y=x+1>0$
$$\begin{cases} x+y=2 \\ y=x+1 \end{cases}$$
Then,  we need

*

*If $-1≤x<0, ~y≥0 $
$$\begin{cases} y-x=2 \\ y=x+1 \end{cases}$$

*

*If $x<-1, ~y<0$
$$\begin{cases} -x-y=2 \\ y=x+1 \end{cases}$$
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