How to draw graph of $|x+y-1| + |2x + y + 1|=1$? 
How to draw graph of $|x+y-1| + |2x + y + 1|=1$ ?

My attempt
I can operate modulus in equations like $|x-1| + |2x + 1|=1$ but I don't know how to proceed in this question since it contains $2$ variable, $x$ and $y$.

If you know already a bit of linear algebra you may proceed as follows:
You surely can draw $|u|+|v|=1$.
Now, consider the mapping
$$\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$
The inverse mapping (just solve for $\begin{pmatrix}x \\ y\end{pmatrix}$) is
$$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}+\begin{pmatrix}-2 \\ 3\end{pmatrix}$$
Since we are dealing with a linear mapping plus a shift (a so-called affine mapping), you only need to find the images of the corners of $|u|+|v|=1$:


*

*$\begin{pmatrix}1 \\ 0\end{pmatrix},\begin{pmatrix}-1 \\ 0\end{pmatrix},\begin{pmatrix}0 \\ 1\end{pmatrix},\begin{pmatrix}0 \\ -1\end{pmatrix}$
Plugging these into the inverse mapping you get as graph the sides of the parallelogram spanned by
$$\boxed{\begin{pmatrix}-3 \\ 5\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 2\end{pmatrix},\begin{pmatrix}-3 \\ 4\end{pmatrix}}$$
