Value of $z$ in the given system of equations

Question

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.

Answer 1

$$\{x\}+y+\lfloor{z}\rfloor=3.1\tag{1}$$

$$x+\lfloor{y}\rfloor+\{z\}=2.4\tag{2}$$

$$\lfloor{x}\rfloor+\{y\}+z=1.3\tag{3}$$

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

Answer 2

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 $1.3$.