2.1. Solve the system of equations in real numbers

$$
\left\{\begin{array}{l}
\{a\}+[b]+\{c\}=2.9 \\
\{b\}+[c]+\{a\}=5.3 \\
\{c\}+[a]+\{b\}=4.0
\end{array}\right.
$$
2.1. Answer: $a=3.1, b=2.2, c=5.8$ or $a=3.6, b=2.7, c=4.3$.

From the third equality, we get that either $\{b\}=\{c\}=0$, but then the first two equalities give contradictory equalities $\{a\}=0.9=0.3$, or $\{c\}+\{b\}=1,[a]=3$. Adding the first two equalities, we get that $\{a\}=0.6$ or $\{a\}=0.1$. In the first case, we get $\{b\}=0.7,\{c\}=0.3$, from which we easily obtain $a=3.6, b=2.7, c=4.3$. In the second case, $\{b\}=0.2$, $\{c\}=0.8, a=3.1, b=2.2, c=5.8$.