\section*{Problem 3 - 211033}

Determine all ordered pairs \((a ; b)\) of real numbers that satisfy the following system of equations!

\[
\begin{aligned}
{[a]+2 b } & =6.6 \\
{[2 a]+3 b } & =11.9
\end{aligned}
\]

Hint: If \(r\) is a real number, \([r]\) denotes the integer \(g\) such that \(g \leq r < g+1\).

For example, \([4.01]=4\), since \(4 \leq 4.01 < 5\); \([7]=7\), since \(7 \leq 7 < 8\); \([-\pi]=-4\), since \(-4 \leq -\pi < -3\).
}

Since \([a] \in \mathbb{Z}\), according to the first equation, the fractional part of \(b\) is either 0.3 or 0.8, because only for these values does \(2b\) have a fractional part of 0.6. However, only the first possibility does not lead to a contradiction with the second equation, because the fractional part of \(3b\), and thus also of \(11.9 = [2a] + 3b\), would be 0.4 in the second case due to \(3 \cdot 0.8 = 2.4\), which is a contradiction.

Therefore, \(b = [b] + 0.3\) and the two equations become \([a] + 2[b] = 6\) and \([2a] + 3[b] = 11\).

We distinguish two cases based on the fractional part of \(a\):

1. Case: \(0 \leq a - [a] < \frac{1}{2}\). Then \(0 \leq 2a - 2[a] < 1\), so \(2[a] \leq 2a < 2[a] + 1\) and thus \([2a] = 2[a]\). This results in the linear system of equations

\[
[a] + 2[b] = 6 \quad ; \quad 2[a] + 3[b] = 11
\]

which has the solution \(([a]; [b]) = (4; 1)\). Therefore, for all \(0 \leq c < \frac{1}{2}\), the pairs \((a; b) = (4 + c; 1.3)\) are solutions to the original system.

2. Case: \(\frac{1}{2} \leq a - [a] < 1\). Then \(1 \leq 2a - 2[a] < 2\), so \(2[a] + 1 \leq 2a < 2[a] + 2\) and thus \([2a] = 2[a] + 1\). This results in the linear system of equations

\[
[a] + 2[b] = 6 \quad ; \quad 2[a] + 1 + 3[b] = 11
\]

which has the solution \(([a]; [b]) = (2; 2)\). Therefore, for all \(0 \leq c < \frac{1}{2}\), the pairs \((a; b) = (2.5 + c; 2.3)\) are solutions to the original system.

In summary, the solution set is:

\[
\left\{(a, b) \left\lvert\, \exists 0 \leq c \leq \frac{1}{2}\right.:(a=4+c \wedge b=1.3) \vee(a=2.5+c \wedge b=2.3)\right\}
\]