2015
Problem - 2666
How many solutions does the following system have?
$$
\left\{
\begin{array}{ll}
\lfloor x \rfloor + 2y &= 1\\
\lfloor y \rfloor + x &=2
\end{array}
\right.
$$
Where $\lfloor x \rfloor$ and $\lfloor y \rfloor$ denote the largest integers not exceeding $x$ and $y$, respectively.
From the $2^{nd}$ equation, we find $x$ must be an integer. From the $1^{st}$ equation, we find $y$ must be either an integer or in a form of $y=\lfloor{x}\rfloor+0.5$
If $y$ is an integer, then
$$
\left\{
\begin{array}{ll}
x + 2y &= 1\\
y + x &=2
\end{array}
\right.
\implies (x, y)=(3, -1)
$$
If $y=\lfloor{y}\rfloor + 0.5$, then
$$
\left\{
\begin{array}{ll}
x + 2\lfloor{y}\rfloor + 1 &= 1\\
\lfloor{y}\rfloor + x &=2
\end{array}
\right.
\implies (x, \lfloor{y}\rfloor)=(4, -2)\implies (x, y)=(4, -1.5)
$$
$$\therefore\quad (x, y) = \boxed{(4, -1.5), (3, -1)}$$