<s>[INST] 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 ?
 [/INST] 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:
$$\mathscr{A}=2\cdot\frac{9}{8}+2\cdot\frac{15}{8}=6\tag3$$
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