## Absolute Value Function

What is the area of the region bounded by the graph of $|x+y|+|x-y|=4$?
Source: NCTM Mathematics Teacher, October 2006

Solution
If $x+y>0$ and $x-y>0$
$|x+y|+|x-y|=4$
$(x+y)+(x-y)=4$
$2x=4$
$x=2$

If $|x+y|<0$ and $|x-y|<0$
$|x+y|+|x-y|=4$
$-(x+y)-(x-y)=4$
$-x-y-x+y=4$
$-2x=4$
$x=-2$

If $|x+y|>0$ and $|x-y|<0$
$|x+y|+|x-y|=4$
$(x+y)-(x-y)=4$
$x+y-x+y=4$
$2y=4$
$y=2$

If $|x+y|<0$ and $|x-y|>0$
$|x+y|+|x-y|=4$
$-(x+y)+(x-y)=4$
$-x-y+x-y=4$
$-2y=4$
$y=-2$

The four graphs intersect and form a square of side length $4$.

The are of the square equals $16$.

Solution: $16$ square units