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.
 A: 
Image courtesy $-$ Martund.
$|x|+|y|=2$ is square on graph paper (you will get it by making cases).
and $y=x+1$ is line.
This is the shortest (not only) method I know.
A: We know that $$|x-y|\le|x|+|y|$$
where equality holds iff $x$ and $y$ are of opposite sign. In this case we know that equality doesn't hold, so $x$ and $y$ are of same sign, so the first equation becomes $|x+y|=2$ and second is $y-x=1$. Making cases for positive and negative sign in first equation, we get the required solutions:
$$\left(\frac12,\frac32\right)\quad\text{and}\quad\left(-\frac32,-\frac12\right)$$
A: 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 $x<-1, ~y<0$
$$\begin{cases} -x-y=2 \\ y=x+1 \end{cases}$$
A: One nice way to think about it is to see |x-a| as the distance between x and a. Visually, it's pretty clear only x=-1.5 and x=0.5 are solutions.
A: You only consider integers, hence you might left out non-integer solutions.
We have $$|x|+|x+1|=2$$
$$|x-0|+|x-(-1)|=2$$
So the solution for $x$ is when the sum of distance from $0$ and the distance from $-1$ is exactly $2$.
Clearly, the point can't be inside $[-1, 0]$.
If $x$ is positive, then we have $2x+1=2$, hence $x=\frac12$.
Similarly, if $x<-1$, by symmetry, $x=-1-\frac12=-\frac32$.
