Question Number 212479 by hardmath last updated on 14/Oct/24
$$\begin{cases}{\mathrm{2x}\:+\:\mathrm{3y}\:−\:\mathrm{z}\:=\:\mathrm{7}}\\{\mathrm{4x}\:−\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{1}}\\{−\mathrm{x}\:+\:\mathrm{5y}\:+\:\mathrm{3z}\:=\:\mathrm{14}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\:\mathrm{x},\mathrm{y},\mathrm{z}\:=\:? \\ $$
Answered by A5T last updated on 14/Oct/24
$$\mathrm{3}\left({ii}\right)+\left({i}\right)\Rightarrow\mathrm{14}{x}+\mathrm{5}{z}=\mathrm{10}…\left({iv}\right) \\ $$$$\mathrm{5}\left({ii}\right)+\left({iii}\right)\Rightarrow\mathrm{19}{x}+\mathrm{13}{z}=\mathrm{19}…\left({v}\right) \\ $$$$\mathrm{13}\left({iv}\right)−\mathrm{5}\left({v}\right)\Rightarrow\mathrm{87}{x}=\mathrm{35}\Rightarrow{x}=\frac{\mathrm{35}}{\mathrm{87}}\Rightarrow{z}=\frac{\mathrm{76}}{\mathrm{87}}\Rightarrow{y}=\frac{\mathrm{205}}{\mathrm{87}} \\ $$
Commented by hardmath last updated on 14/Oct/24
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$