$ 0\le {a}+{c} < 2$ means that $ 0 \le 2.9-[b] < 2$ or, $ 0.9 < [b] \le 2.9$, so $ [b]$ can equal either $ 1$ or $ 2$. Similarly, we find that $ 3.3 < [c] \le 5.3$ so $ [c]$ can equal $ 4$ or $ 5$. And, $ 2 < [a] \le 4$ so $ [a]$ can equal $ 3$ or $ 4$. Try each possible case. It's long, but somebody has to do it.

just there are fewer cases just because from: $ \{c\} + [a] + \{b\} = 4.0$ $ \{c\} = \{b\}=0$ with no solution for $ \{a\}$ or $ 0 < \{c\} + \{b\} <2$

From the third equation we find that $ \{b\}+\{c\}$ is an integer Moreover adding the first two expressions and combining with the previous result we obtain $ 2\{a\}=0.2$ 1)$ \{a\}=0.1$ and 2)$ \{a\}=0.6$ 1)Replacing in the second equation: $ \{b\}=0.2$ and doing the same in the third equation $ \{c\}=0.8$ $ [a]=3$ $ [b]=2$ $ [c]=5$ 2) doing the same as above we find $ \{b\}=0.7$ $ \{c\}=0.3$ and $ [a]=3$ $ [b]=2$ $ [c]=4$ Thus this are the only solutions Daniel

this problem is similar to that problem in the book 'mathematical olympiad challenges' chapter 2.5