3. Denote by $[a]$ the greatest integer less than or equal to $a$. Let $N$ be an integer, $x$ and $y$ be numbers satisfying the simultaneous equations $\left\{\begin{array}{l}{[x]+2 y=N+2} \\ {[y]+2 x=3-N}\end{array}\right.$. Find $x$ in terms of $N$.
(1 mark)我們把小於或等於 $a$ 的最大整數記作 $[a]$ 。設 $N$ 為整數, 且 $x$ 和 $y$ 滿足聯立方程 $\left\{\begin{array}{l}{[x]+2 y=N+2} \\ {[y]+2 x=3-N}\end{array}\right.$ 。求 $x$, 答案以 $N$ 表示。
3. $\frac{3}{2}-N$
3. From the first equation, we have $2 y=N+2-[x]$, which is an integer. Hence $y$ is either an integer or midway between two integers. The same is true for $x$ by looking at the second equation. Hence, either $[x]=x$ or $[x]=x-\frac{1}{2}$, and either $[y]=y$ or $[y]=y-\frac{1}{2}$.
If $x$ and $y$ are both integers, $[x]=x$ and $[y]=y$. Solving the equations, we get $x=\frac{4}{3}-N$ and $y=N+\frac{1}{3}$ which are not consistent with our original assumptions.
If $x$ is an integer and $y$ is midway between two integers, $[x]=x$ and $[y]=y-\frac{1}{2}$. Solving the equations, we have $x=\frac{5}{3}-N$ and $y=N+\frac{1}{6}$ which are not consistent with our original assumptions.
Similarly, if $x$ is midway between two integers and $y$ is an integer, we have $[x]=x-\frac{1}{2}$ and $[y]=y$. Solving the equations, we have $x=\frac{7}{6}-N$ and $y=N+\frac{2}{3}$ which are again not consistent with our original assumptions.
Finally, if $x$ and $y$ are both midway between two integers, we have $[x]=x-\frac{1}{2}$ and $[y]=y-\frac{1}{2}$. Solving the equations, we have $x=\frac{3}{2}-N$ and $y=N+\frac{1}{2}$. This gives the correct answer.