5. Find the area of the region of the $x y$-plane defined by the inequality $|x|+|y|+|x+y| \leq 1$.
Solution:
$3 / 4$

To graph this region we divide the $x y$-plane into six sectors depending on which of $x, y, x+y$ are $\geq 0$, or $\leq 0$. The inequality simplifies in each case:
\begin{tabular}{c|c|c} 
Sector & Inequality & Simplified inequality \\
\hline$x \geq 0, y \geq 0, x+y \geq 0$ & $x+y+(x+y) \leq 1$ & $x+y \leq 1 / 2$ \\
$x \geq 0, y \leq 0, x+y \geq 0$ & $x-y+(x+y) \leq 1$ & $x \leq 1 / 2$ \\
$x \geq 0, y \leq 0, x+y \leq 0$ & $x-y-(x+y) \leq 1$ & $y \geq-1 / 2$ \\
$x \leq 0, y \geq 0, x+y \geq 0$ & $-x+y+(x+y) \leq 1$ & $y \leq 1 / 2$ \\
$x \leq 0, y \geq 0, x+y \leq 0$ & $-x+y-(x+y) \leq 1$ & $x \geq-1 / 2$ \\
$x \leq 0, y \leq 0, x+y \leq 0$ & $-x-y-(x+y) \leq 1$ & $x+y \geq-1 / 2$
\end{tabular}

We then draw the region; we get a hexagon as shown. The hexagon intersects each region in an isosceles right triangle of area $1 / 8$, so the total area is $6 \cdot 1 / 8=3 / 4$.