2007
Problem - 3668
How many pairs of ordered real numbers $(x, y)$ are there such that
$$
\left\{
\begin{array}{ccl}
\mid x\mid + y &=& 12\\
x + \mid y \mid &=&6
\end{array}
\right.
$$
If $x\ge 0$, then
$$
\left\{
\begin{array}{ccl}
x + y &=& 12\\
x + \mid y \mid &=&6
\end{array}
\right.
\implies \mid y\mid - y=-6
$$
This is impossible because $\mid y\mid -y\ge 0$.
If $x < 0$, then
$$
\left\{
\begin{array}{ccl}
-x + y &=& 12\\
x + \mid y \mid &=&6
\end{array}
\right.
\implies \mid y\mid + y=18\implies y=9\implies x=-3
$$
In conclusion, there is only one solution to the given system.