Find the area of that portion of the $xy$-plane which is enclosed by the curve with equation
|2x-1|+|2x+1|+\frac{4|y|}{\sqrt{3}}=4.
Solution. The curve is clearly symmetric with respect to both the $x$-axis and the $y$-axis. Our
only weapon from here is to try to perhaps write $y$ in terms of $x$.
For $x<-1/2$, then $|2x-1|=-(2x-1)=1-2x$, $|2x+1|=-(2x+1)$,
and we find that $y=\pm\sqrt{3}(x+1)$. These lines intersect at $x=-1$.
For $-1/2<x<1/2$, we have $|2x-1|=1-2x$, $|2x+1|=2x+1$, and thus $y=\pm\sqrt{3}/2$.
For $x>1/2$, we have $|2x-1|=2x-1$, $|2x+1|=2x+1$, and thus $y=\pm\sqrt{3}(1-x)$. These two lines intersect at $x=1$.
This gives us a regular hexagon of sidelength 1 with vertices
$(\pm 1/2, \sqrt{3}/2)$, $(\pm 1/2, -\sqrt{3}/2)$, and $(\pm 1,0)$. The
enclosed area -- that of six equilateral triangles of unit edge length
is $(6)(1/2)(1)(1)(\sqrt{3}/2)=3\sqrt{3}/2$.