calculation of $(x,y,z)$ in $x+\lfloor y \rfloor -\{z\} = 2.98\;\;,\lfloor x \rfloor +\{y\}-z = 4.05\;\;,-\{x\}+y+\lfloor z \rfloor = 5.01$ The no. of real solution of the equation 
$x+\lfloor y \rfloor -\{z\} = 2.98\;\;,\lfloor x \rfloor +\{y\}-z = 4.05\;\;,-\{x\}+y+\lfloor z \rfloor = 5.01$.
Here I did not understand How can I solve it.
help me
Thanks. 

I'm assuming that indeed $\{a\} = a - \lfloor a\rfloor$. Since then $0 \leqslant \{a\} < 1$ for all $a$, we have
$$-1 < \{a\} - \{b\} < 1$$
for all $a,b$. Then from the given equations we obtain, by writing $x = \lfloor x\rfloor + \{x\}$ and analogously for $y,z$, the three equations
$$\begin{gather}
\{ x\} - \{ z\} = -0.02,\tag{1}\\
\{ y\} - \{ z\} = 0.05,\tag{2}\\
\{ y\} - \{ x\} = 0.01.\tag{3}
\end{gather}$$
Adding $(1)$ and $(3)$ yields $\{y\} - \{z\} = -0.01$, which contradicts $(2)$, hence there is no solution to the system.
