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

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.

• 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. – Tapi May 11 at 9:39
• @Tapi Please write an answer showing only this step that you mention. I'll will fully upvote it. – PranavGupta53535 May 11 at 9:42
• 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.. – Oolong milk tea May 11 at 9:45
• Please do not use images. – Dietrich Burde May 11 at 10:33
• I'll try not to next time. – PranavGupta53535 May 11 at 10:34

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 $$[c]=8$$, $$\{b\}=0.1$$.