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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.

My first step.

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    $\begingroup$ try simultaneously solving two equations on the right at a time. If you subtract and take minus common, you'd get a+b, b+c, and c+a. $\endgroup$ – Tapi May 11 at 9:39
  • $\begingroup$ @Tapi Please write an answer showing only this step that you mention. I'll will fully upvote it. $\endgroup$ – PranavGupta53535 May 11 at 9:42
  • $\begingroup$ You can find the integer part of a,b,c from this above(Don't be afraid of discussion). Here are only 4 cases $(\lfloor a \rfloor,\lfloor b \rfloor,\lfloor c \rfloor)=(12/13,14/15,13/14)$, and their sum must be even. It's not difficult with some patient.. $\endgroup$ – Oolong milk tea May 11 at 9:45
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    $\begingroup$ Please do not use images. $\endgroup$ – Dietrich Burde May 11 at 10:33
  • $\begingroup$ I'll try not to next time. $\endgroup$ – PranavGupta53535 May 11 at 10:34
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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 $[c]=8$, $\{b\}=0.1$.

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