Question Number 212424 by hardmath last updated on 13/Oct/24
$$\begin{cases}{\mathrm{x}\:+\:\mathrm{2y}\:−\:\mathrm{3z}\:=\:\mathrm{1}}\\{\mathrm{2x}\:−\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{4}}\\{\mathrm{3x}\:+\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{7}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\mathrm{x},\mathrm{y},\mathrm{z}\:=\:? \\ $$
Answered by A5T last updated on 13/Oct/24
$$\left({ii}\right)+\left({iii}\right)\Rightarrow\mathrm{5}{x}+\mathrm{3}{z}=\mathrm{11}…\left({iv}\right) \\ $$$$\mathrm{2}\left({iii}\right)−\left({i}\right)\Rightarrow\mathrm{5}{x}+\mathrm{7}{z}=\mathrm{13}…\left({v}\right) \\ $$$$\left({v}\right)−\left({iv}\right)\Rightarrow\mathrm{4}{z}=\mathrm{2}\Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}=\frac{\mathrm{19}}{\mathrm{10}}\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{1}.\mathrm{9},\mathrm{0}.\mathrm{3},\mathrm{0}.\mathrm{5}\right) \\ $$
Commented by hardmath last updated on 13/Oct/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$