Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$ 
Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$

I tried taking four cases which are:
$$x+3y \geq 0, x-y \geq 0$$
$$x+3y \geq 0, x-y \leq 0$$
$$x+3y \leq 0, x-y \geq 0$$
$$x+3y \leq 0, x-y \leq 0$$
But it becomes so confusing to draw the region. Any better way ?
 A: The transformation
$$ 
\begin{align}
u &= x + y \\
v &= 2y
\end{align}
$$
transforms the region
$$
2 \leq|x+3 y|+|x-y| \leq 4
$$
to
$$
 2 \leq|u+v|+|u-v| \leq 4
$$
and increases the area by a factor of two (why?). The new region is limited by two squares with side length $2$ and $4$, respectively (compare How to solve $|a+b|+|a-b|=c$?),  so that its area can be computed easily.
A: Here is how I would resolve the absolute equations.
$2 \leq|x+3 y|+|x-y| \leq 4$
If $x \geq y,$
$2 \leq|x+3 y|+ x-y \leq 4$
For $x+3y \geq 0, x+3y \leq 0,$ we get
$1 \leq x+y \leq 2, - 1 \leq y \leq -\frac{1}{2}$
Similarly for $x \leq y$, we have
$-2 \leq x+y \leq -1, \frac{1}{2} \leq y \leq 1$
That gives you the below region to find area of,

Finding area may be easier with a transformation and the transformation choice is obvious but given the focus of the question, I am not getting there.
A: Well, if you're looking for a plot:

Now, if we want to find the area of the part where $y\ge0$ and $x\ge0$, we can take a look at the following picture:

We can find the shaded area by finding:
$$3\cdot\left(\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}\right)+\frac{1}{2}\cdot\frac{1}{2}+\int_1^2\left(2-x\right)\space\text{d}x=\frac{9}{8}\tag1$$
And the place where $y\ge0$ and $x\le0$, we can take a look at the following picture:

We can find the shaded area by finding:
$$3\cdot\frac{1}{2}\cdot1+\frac{1}{2}\cdot\frac{1}{2}+\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}=\frac{15}{8}\tag 2$$
So, the total area is given by:
$$\mathscr{A}=2\cdot\frac{9}{8}+2\cdot\frac{15}{8}=6\tag3$$
