Question Number 120140 by help last updated on 29/Oct/20
Answered by nimnim last updated on 29/Oct/20
$$\left({x},{y},{z}\right)=\left(\mathrm{2},\mathrm{4},−\mathrm{3}\right) \\ $$
Commented by help last updated on 29/Oct/20
$${solution}\:{please} \\ $$
Answered by nimnim last updated on 29/Oct/20
$${Breaking}\:{the}\:{brackets},\:{we}\:{get} \\ $$$$\mathrm{5}{x}+\mathrm{10}{y}−\mathrm{12}{x}−\mathrm{16}{z}−\mathrm{2}{x}−\mathrm{6}{y}+\mathrm{10}{z}=\mathrm{16} \\ $$$$−\mathrm{9}{x}+\mathrm{4}{y}−\mathrm{6}{z}=\mathrm{16}.............\left({i}\right) \\ $$$$\:\:\:\mathrm{6}{x}−\mathrm{2}{y}+\mathrm{3}{x}−\mathrm{6}{z}+\mathrm{8}{x}−\mathrm{12}{y}+\mathrm{4}{z}=−\mathrm{16} \\ $$$$\mathrm{17}{x}−\mathrm{14}{y}−\mathrm{2}{z}=−\mathrm{16}........\left({ii}\right) \\ $$$$\mathrm{4}{y}−\mathrm{8}{z}+\mathrm{4}{x}−\mathrm{8}{y}−\mathrm{6}−\mathrm{3}{x}−\mathrm{12}{y}+\mathrm{6}{z}=\mathrm{62} \\ $$$$\:\:\:\:{x}−\mathrm{16}{y}−\mathrm{2}{z}=−\mathrm{56}........\left({iii}\right) \\ $$$$ \\ $$$$\left({ii}\right)−\left({iii}\right)\Rightarrow\mathrm{16}{x}+\mathrm{2}{y}=\mathrm{40}\Rightarrow\mathrm{8}{x}+{y}=\mathrm{20}.....\left({iv}\right) \\ $$$$\left({i}\right)−\mathrm{3}\left({iii}\right)\Rightarrow−\mathrm{12}{x}+\mathrm{52}{y}=\mathrm{184}\Rightarrow−\mathrm{3}{x}+\mathrm{13}{y}=\mathrm{46}....\left({v}\right) \\ $$$${from}\:\left({iv}\right){and}\:\left({v}\right) \\ $$$${x}=\mathrm{2}\:{and}\:{y}=\mathrm{4} \\ $$$$\left({iii}\right)\Rightarrow\mathrm{2}−\mathrm{16}\left(\mathrm{4}\right)−\mathrm{2}{z}=−\mathrm{56} \\ $$$$\Rightarrow{z}=−\mathrm{3} \\ $$$$ \\ $$