Question Number 188536 by normans last updated on 03/Mar/23
$$ \\ $$$$\:\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}; \\ $$$$\:\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\mathrm{5}}\\{\boldsymbol{{z}}\:−\:\boldsymbol{{yx}}\:=\:\mathrm{7}}\\{\boldsymbol{{z}}\:=\:\mathrm{1}\:+\:\boldsymbol{{x}}}\end{matrix}\right\}\:\:\boldsymbol{{x}};\boldsymbol{{y}};\boldsymbol{{z}}\:=??\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 03/Mar/23
$$\:\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\mathrm{5}…\left({i}\right)}\\{\boldsymbol{{z}}\:−\:\boldsymbol{{yx}}\:=\:\mathrm{7}…\left({ii}\right)}\\{\boldsymbol{{z}}\:=\:\mathrm{1}\:+\:\boldsymbol{{x}}….\left({iii}\right)}\end{matrix}\right\}\:\:\boldsymbol{{x}};\boldsymbol{{y}};\boldsymbol{{z}}\:=?? \\ $$$$\:\left({i}\right)\Rightarrow{x}+{y}−\left(\mathrm{1}+{x}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:{y}=\mathrm{6} \\ $$$$\left({ii}\right)\Rightarrow{z}−\mathrm{6}{x}=\mathrm{7} \\ $$$$\left({iii}\right)\Rightarrow{z}−{x}=\mathrm{1} \\ $$$$\:\:\:\:\:\:−\mathrm{5}{x}=\mathrm{6}\Rightarrow{x}=−\mathrm{6}/\mathrm{5} \\ $$$$\:{z}=\mathrm{1}+{x}=\mathrm{1}−\frac{\mathrm{6}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\left({x},{y},{z}\right)=\left(−\frac{\mathrm{6}}{\mathrm{5}},\mathrm{6},−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$
Answered by Yhusuph last updated on 03/Mar/23
$${x}+{y}−\mathrm{7}−{xy}=\mathrm{5} \\ $$$${x}+{y}−{xy}=\mathrm{13}……\left({i}\right) \\ $$$${x}=\frac{\mathrm{13}−{y}}{\mathrm{1}−{y}}……\left({ii}\right) \\ $$$$\mathrm{7}+{xy}=\mathrm{1}+{x} \\ $$$${x}−{xy}=\mathrm{6}……\left({iii}\right) \\ $$$${x}\left(\mathrm{1}−{y}\right)=\mathrm{6} \\ $$$$\left(\frac{\mathrm{13}−{y}}{\mathrm{1}−{y}}\right)\left(\mathrm{1}−{y}\right)=\mathrm{6} \\ $$$$\mathrm{13}−{y}=\mathrm{6} \\ $$$${y}=\mathrm{7} \\ $$$${x}=\frac{\mathrm{6}}{\mathrm{1}−{y}}=\frac{\mathrm{6}}{\mathrm{1}−\mathrm{7}}=−\mathrm{1} \\ $$$${z}=\:\mathrm{1}+{x}\:=\:\mathrm{1}+\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\therefore\:{x},{y\&z}\:{are}\:−\mathrm{1},\mathrm{7}\:\&\:\mathrm{0}\:{respectively} \\ $$$$ \\ $$$$\partial\epsilon\eta{k}\epsilon\eta\:{la}\int{t}\:\beta\theta{r}\eta \\ $$$$ \\ $$$$ \\ $$