Question Number 213430 by hardmath last updated on 05/Nov/24
![x,y,z ∈ R { ((x + [y] + {z} = 9,4)),(([x] + {y} + z = 11,3)),(({x} + y + [z] = 10,5)) :} find: x = ?](https://www.tinkutara.com/question/Q213430.png)
$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\left[\mathrm{y}\right]\:+\:\left\{\mathrm{z}\right\}\:=\:\mathrm{9},\mathrm{4}}\\{\left[\mathrm{x}\right]\:+\:\left\{\mathrm{y}\right\}\:+\:\mathrm{z}\:=\:\mathrm{11},\mathrm{3}}\\{\left\{\mathrm{x}\right\}\:+\:\mathrm{y}\:+\:\left[\mathrm{z}\right]\:=\:\mathrm{10},\mathrm{5}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by A5T last updated on 05/Nov/24
![(i)−(ii): {x}+[y]−{y}−[z]=−1.9...(iv) (iv)−(iii)⇒[y]−{y}−y−2[z]=−12.4 ⇒{y}+[z]=6.2; ⇒[z]=6⇒{y}=0.2 (iii)⇒{x}+[y]+{y}=4.5⇒{x}+[y]=4.3 ⇒[y]=4⇒{x}=0.3 (i)⇒[x]+0.3+4+{z}=9.4⇒[x]+{z}=5.1 ⇒[x]=5;{z}=0.1⇒x=5.3⇒(x,y,z)=(5.3,4.2,6.1)](https://www.tinkutara.com/question/Q213434.png)
$$\left({i}\right)−\left({ii}\right):\:\left\{{x}\right\}+\left[{y}\right]−\left\{{y}\right\}−\left[{z}\right]=−\mathrm{1}.\mathrm{9}…\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right)\Rightarrow\left[{y}\right]−\left\{{y}\right\}−{y}−\mathrm{2}\left[{z}\right]=−\mathrm{12}.\mathrm{4} \\ $$$$\Rightarrow\left\{{y}\right\}+\left[{z}\right]=\mathrm{6}.\mathrm{2};\:\Rightarrow\left[{z}\right]=\mathrm{6}\Rightarrow\left\{{y}\right\}=\mathrm{0}.\mathrm{2} \\ $$$$\left({iii}\right)\Rightarrow\left\{{x}\right\}+\left[{y}\right]+\left\{{y}\right\}=\mathrm{4}.\mathrm{5}\Rightarrow\left\{{x}\right\}+\left[{y}\right]=\mathrm{4}.\mathrm{3} \\ $$$$\Rightarrow\left[{y}\right]=\mathrm{4}\Rightarrow\left\{{x}\right\}=\mathrm{0}.\mathrm{3} \\ $$$$\left({i}\right)\Rightarrow\left[{x}\right]+\mathrm{0}.\mathrm{3}+\mathrm{4}+\left\{{z}\right\}=\mathrm{9}.\mathrm{4}\Rightarrow\left[{x}\right]+\left\{{z}\right\}=\mathrm{5}.\mathrm{1} \\ $$$$\Rightarrow\left[{x}\right]=\mathrm{5};\left\{{z}\right\}=\mathrm{0}.\mathrm{1}\Rightarrow{x}=\mathrm{5}.\mathrm{3}\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{5}.\mathrm{3},\mathrm{4}.\mathrm{2},\mathrm{6}.\mathrm{1}\right) \\ $$