Solve $x + \lfloor y \rfloor + \{ z \} = 13.2$; $\{ x \} + y + \lfloor z \rfloor = 15.1$; $\lfloor x \rfloor + \{ y \} + z = 14.3$

Question

The equation system is —

$x + \lfloor y \rfloor + \{ z \} = 13.2$

$\{ x \} + y + \lfloor z \rfloor = 15.1$

$\lfloor x \rfloor + \{ y \} + z = 14.3$

Now I've tried substituting $n$ with $\lfloor n \rfloor + \{ n \}$ everywhere possible and then gone on with algebraic manipulations. But everything gets messy from there. I tried solving the problem more than thrice over the past few days, but always ended up with different answers.

Answer 1

Hint:

Add them up, $a+b+c=\dfrac{13.2+15.1+14.3}{2}=21.3$.

So, $\{b\}+[c]=21.3-13.2=8.1$ and hence $[c]=8$, $\{b\}=0.1$.