Question Number 193284 by universe last updated on 09/Jun/23
$$\:\:\:\mathrm{6}{y}\:−\mathrm{2}{xy}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\mathrm{8}{z}\:−\:{yz}\:=\:\mathrm{9} \\ $$$$\:\:\:\mathrm{10}{x}\:−\:\mathrm{4}{xz}\:=\:\mathrm{8}\: \\ $$$${find}\:{x}+{y}\:+{z}\:=\:? \\ $$
Answered by AST last updated on 09/Jun/23
$${x}=\mathrm{3}−\frac{\mathrm{4}}{\mathrm{2}{y}};{y}=\frac{\mathrm{8}{z}−\mathrm{9}}{{z}}=\mathrm{8}−\frac{\mathrm{9}}{{z}};{z}=\frac{\mathrm{10}{x}−\mathrm{8}}{\mathrm{4}{x}}=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{2}}{{x}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{2}}{\frac{\mathrm{6}{y}−\mathrm{4}}{\mathrm{2}{y}}}=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{4}{y}}{\mathrm{6}{y}−\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{8}−{y}} \\ $$$$\Rightarrow\frac{\mathrm{15}{y}−\mathrm{10}−\mathrm{4}{y}}{\mathrm{6}{y}−\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{8}−{y}}\Rightarrow\left(\mathrm{11}{y}−\mathrm{10}\right)\left(\mathrm{8}−{y}\right)=\mathrm{54}{y}−\mathrm{36} \\ $$$$\Rightarrow\mathrm{11}{y}^{\mathrm{2}} −\mathrm{44}{y}+\mathrm{44}=\mathrm{0}\Rightarrow{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4}=\mathrm{0}\Rightarrow{y}=\mathrm{2}\left({twice}\right) \\ $$$${y}=\mathrm{2}\Rightarrow{z}=\frac{\mathrm{9}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{x}=\mathrm{2}\Rightarrow{x}+{y}+{z}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$