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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.

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

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

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

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

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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 $1.3$.

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  • $\begingroup$ I am not getting your hint $\endgroup$ – Archis Welankar Jun 27 '16 at 14:36
  • $\begingroup$ @ArchisWelankar: It's not a hint; it's an argument that your system of equations is unsolvable. $\endgroup$ – hmakholm left over Monica Jun 27 '16 at 14:41
  • $\begingroup$ No I have two options in my question $2.2,4.3$ how to see which is correct? $\endgroup$ – Archis Welankar Jun 27 '16 at 15:23
  • $\begingroup$ @ArchisWelankar: As I explain, there is no solution. This means that $2.2$ is not correct, and $4.3$ is not correct, and $117$ is not correct, and no other number you can think of is correct either. You don't have to believe me, of course, but if you insist that my argument is wrong, it would be polite for you to explain what you feel is unconvincing about it. $\endgroup$ – hmakholm left over Monica Jun 27 '16 at 15:36
  • $\begingroup$ I never said like that but this question was given in my test of mcq out of which one was correct also there was no option Like no solution so I am confused $\endgroup$ – Archis Welankar Jun 27 '16 at 15:42

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