, 2009

Solution:
The first equation is equivalent to
$$|x|+|y|=1$$
or
$$|x|+|y|=3$$
The graph of the first equation is symmetric with respect to both axes. In the first quadrant it is reduced to $x+y=1$, whose graph is segment connecting points $(1,0)$ and $(0,1)$. Thus, the graph of
$$|x|+|y|=1$$
is square with vertices $(1,0),(0,1),(-1,0)$ and $(0,-1)$. Similarly, the graph of
$$|x|+|y|=3$$
is a square with vertices $(3,0),(0,3),(-3,0)$ and $(0,-3)$. The graph of the second equation of the system is a straight line with slope $a$ passing through $(0,5)$. This line intersects the graph of the first equation in three points exactly, when passing through one of the points $(1,0)$ or $(-1,0)$. This happens if and only if $a=5$ or $a=-5$.