Let $n$ be the real number. Define $[n]$ be the integer part of $n$, and $\{n\}$ be the decimal part of $n$. Solve the following equations:
$$\begin{array}{ccccccc}
\{x\} & + & [y] & + & \{z\} & = 2.9\\
\{y\} & + & [z] & + & \{x\} & = 5.3\\
\{z\} & + & [x] & + & \{y\} & = 4.0\\
\end{array}$$
Solution: Because $\{z\}+[x]+\{y\}=4.0$, $\{z\}+\{y\}$ is either $0.0$ or $1.0$.
If $\{z\}+\{y\}=0.0$, then $\{y\}=\{z\}=0.0$. This would result in $\{x\}=0.9$ and $\{x\}=0.3$, which is impossible for ${x}$ to take two values at the same time. Therefore $\{z\}+\{y\}=1.0$.
Because $\{x\}+[y]+\{z\}=2.9$, we have $\{x\}+\{z\}=0.9$ or $\{x\}+\{z\}=1.9$
If $\{x\}+\{z\}=1.9$, therefore $\{x\}>0.9$, $\{z\}>0.9$. Therefore $\{y\}+\{x\}=1.3$. We have:
$$\begin{array}{ccccccc}
\{x\} & + & \{z\} & = 1.9\\
\{y\} & + & \{x\} & = 1.3\\
\{z\} & + & \{y\} & = 1.0\\
\end{array}$$
Solving the above equations, we have $\{x\}=1.1$, $\{y\}=0.2$, and $\{z\}=0.8$, which is invalid solution as $\{x\}\ge 1.0$. Therefore $\{x\}+\{z\}=0.9$.
Therefore, $\{x\}+\{z\}$ is either $0.3$ or $1.3$.
If $\{x\}+\{z\}=0.3$, we have
$$\begin{array}{ccccccc}
\{x\} & + & \{z\} & = 0.9\\
\{y\} & + & \{x\} & = 0.3\\
\{z\} & + & \{y\} & = 1.0\\
\end{array}$$
Solving the above equations, we have $\{x\}=0.1$, $\{y\}=0.2$, $\{z\}=0.8$. This results in $x=3.1$, $y=2.2$, and $z=5.8$.
If $\{x\}+\{z\}=1.3$, we have
$$\begin{array}{ccccccc}
\{x\} & + & \{z\} & = 0.9\\
\{y\} & + & \{x\} & = 1.3\\
\{z\} & + & \{y\} & = 1.0\\
\end{array}$$
Solving the above equations, we have $\{x\}=0.6$, $\{y\}=0.7$, $\{z\}=0.3$. This results in $x=3.6$, $y=2.7$, and $z=4.3$.
In conclusion there are two solutions for $(x, y, z)$, i.e. $(3.1, 2.2, 5.8)$ and $(3.6, 2.7, 4.3)$.
