Value of $z$ in the given system of equations If $$\{x\}+y+\lfloor{z}\rfloor=3.1$$
$$x+\lfloor{y}\rfloor+\{z\}=2.4$$
$$\lfloor{x}\rfloor+\{y\}+z=1.3$$
then find the value of $z$.
My attempt:
I  converted fractional part of every equation to greatest integer to get $$x+y+z=3.4$$ 
but I don't see any way to solve it.
 A: Note: This answers an earlier version of the question before the $\{\;\}$ brackets were added in the equations.
IF $[x]$ means the integer prat of $x$, then you can rewrite your equations as
$$ x+y+z-\{z\} = 3.1 \\ x+y+z - \{y\} = 2.4 \\ x+y+z-\{x\} = 1.3 $$
But it's easy to see that this can't have a solution at all, because $x+y+z$ must be at least $3.1$ for the first equation to be true -- but then $x+y+z-\{x\}$ cannot be smaller then $2.1$. In particular, it is then impossible for it to be $1.3$.
A: $$\{x\}+y+\lfloor{z}\rfloor=3.1\tag{1}$$
$$x+\lfloor{y}\rfloor+\{z\}=2.4\tag{2}$$
$$\lfloor{x}\rfloor+\{y\}+z=1.3\tag{3}$$
Observe that $$\{a\}+\lfloor{a}\rfloor=a$$
Now, add $(1),(2)$ and $(3)$, to get
$$x+y+z=3.4\tag{4}$$
Then, 
Subtract $(1)$ from $(4)$ to get
$$\lfloor{x}\rfloor+\{z\}=0.3\tag{5}$$
Similarly, 
Subtract $(2)$ from $(4)$ to get
$$\{y\}+\lfloor{z}\rfloor=1.0\tag{6}$$
Subtract $(3)$ from $(4)$ to get
$$\{x\}+\lfloor{y}\rfloor=2.1\tag{7}$$
As $0 \leq \{a\} <1$


*

*From $(5)$, we get $\lfloor{x}\rfloor=0, \{z\}=0.3$

*From $(6)$, we get $\{y\}=0,\lfloor{z}\rfloor=1$

*From $(7)$, we get $\{x\}=0.1, \lfloor{y}\rfloor=2$



Thus, $(x,y,z)=(0.1,2.0,1.3)$

