
Problem 9.1. a) Draw all points in the plane with coordinates $(x ; y)$ such that

$$
(|3 x-y|-3)(|3 x+y|-3)=0
$$

b) Find all $x$ and $y$ for which

$$
\begin{array}{ccc}
(|3 x-y|-3)(|3 x+y|-3) & = & 0 \\
y-\{4 x\} & = & 0 \\
-1 \leq x \leq 1 & &
\end{array}
$$

(For a real number $x$ we denote the unique number in the interval $[0,1)$ for which $x-\{x\}$ is an integer by $\{x\}$ ).


Solution: a) It is easy to see that these are the points on the four lines $l_{1}: 3 x-y=3, l_{2}: 3 x-y=-3, l_{3}: 3 x+y=3$ and $l_{4}: 3 x+y=$ -3 .

b) There are two types of solutions:

1. Solutions of $\left\lvert\, \begin{array}{ccc}y & = & -3 x+3 \\ y & = & \{4 x\} \\ -1 & \leq x \leq & 1\end{array}\right.$
2. Solutions of $\left\lvert\, \begin{array}{ccc}y & = & 3 x+3 \\ y & = & \{4 x\} \\ -1 & \leq x \leq & 1\end{array}\right.$

Denote the integer part of $4 x$ by $[4 x]$, i.e. $4 x=[4 x]+\{4 x\}$. In the first case we have $-3 x+3=\{4 x\}=4 x-[4 x]$, i.e $7 x=3+[4 x]$.

Since the right hand side is an integer all posiible values of $x$ are $0, \pm \frac{1}{7}, \pm \frac{2}{7}, \ldots, \pm 1$. Direct verification shows that only $x=\frac{5}{7}, y=\frac{6}{7}$ $x=\frac{6}{7}, y=\frac{3}{7} ; x=1, y=0$ are solutions.

In the second case $3 x+3=\{4 x\}=4 x-[4 x]$, i.e. $x=3+[4 x]$. It follows that $x$ is an integer, i.e. $x=-1,0$ or 1 . Direct verification shows that solution is only $x=-1, y=0$.

Thus, there are four solutions:

$$
x=\frac{5}{7}, y=\frac{6}{7} ; x=\frac{6}{7}, y=\frac{3}{7} ; x=1, y=0 ; x=-1, y=0
$$
