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Question Number 188536 by normans last updated on 03/Mar/23

solve the equation; {: ((x + y − z =5)),((z − yx = 7)),((z = 1 + x)) } x;y;z =??

$$ \\ $$$$\:\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}}; \\ $$$$\:\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\mathrm{5}}\\{\boldsymbol{{z}}\:−\:\boldsymbol{{yx}}\:=\:\mathrm{7}}\\{\boldsymbol{{z}}\:=\:\mathrm{1}\:+\:\boldsymbol{{x}}}\end{matrix}\right\}\:\:\boldsymbol{{x}};\boldsymbol{{y}};\boldsymbol{{z}}\:=??\:\:\:\:\:\: \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 03/Mar/23

{: ((x + y − z =5...(i))),((z − yx = 7...(ii))),((z = 1 + x....(iii))) } x;y;z =?? (i)⇒x+y−(1+x)=5 ⇒ y=6 (ii)⇒z−6x=7 (iii)⇒z−x=1 −5x=6⇒x=−6/5 z=1+x=1−(6/5)=−(1/5) (x,y,z)=(−(6/5),6,−(1/5))

$$\:\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\mathrm{5}...\left({i}\right)}\\{\boldsymbol{{z}}\:−\:\boldsymbol{{yx}}\:=\:\mathrm{7}...\left({ii}\right)}\\{\boldsymbol{{z}}\:=\:\mathrm{1}\:+\:\boldsymbol{{x}}....\left({iii}\right)}\end{matrix}\right\}\:\:\boldsymbol{{x}};\boldsymbol{{y}};\boldsymbol{{z}}\:=?? \\ $$$$\:\left({i}\right)\Rightarrow{x}+{y}−\left(\mathrm{1}+{x}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:{y}=\mathrm{6} \\ $$$$\left({ii}\right)\Rightarrow{z}−\mathrm{6}{x}=\mathrm{7} \\ $$$$\left({iii}\right)\Rightarrow{z}−{x}=\mathrm{1} \\ $$$$\:\:\:\:\:\:−\mathrm{5}{x}=\mathrm{6}\Rightarrow{x}=−\mathrm{6}/\mathrm{5} \\ $$$$\:{z}=\mathrm{1}+{x}=\mathrm{1}−\frac{\mathrm{6}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\left({x},{y},{z}\right)=\left(−\frac{\mathrm{6}}{\mathrm{5}},\mathrm{6},−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$

Answered by Yhusuph last updated on 03/Mar/23

x+y−7−xy=5 x+y−xy=13......(i) x=((13−y)/(1−y))......(ii) 7+xy=1+x x−xy=6......(iii) x(1−y)=6 (((13−y)/(1−y)))(1−y)=6 13−y=6 y=7 x=(6/(1−y))=(6/(1−7))=−1 z= 1+x = 1+(−1)=0 ∴ x,y&z are −1,7 & 0 respectively ∂εηkεη la∫t βθrη

$${x}+{y}−\mathrm{7}−{xy}=\mathrm{5} \\ $$$${x}+{y}−{xy}=\mathrm{13}......\left({i}\right) \\ $$$${x}=\frac{\mathrm{13}−{y}}{\mathrm{1}−{y}}......\left({ii}\right) \\ $$$$\mathrm{7}+{xy}=\mathrm{1}+{x} \\ $$$${x}−{xy}=\mathrm{6}......\left({iii}\right) \\ $$$${x}\left(\mathrm{1}−{y}\right)=\mathrm{6} \\ $$$$\left(\frac{\mathrm{13}−{y}}{\mathrm{1}−{y}}\right)\left(\mathrm{1}−{y}\right)=\mathrm{6} \\ $$$$\mathrm{13}−{y}=\mathrm{6} \\ $$$${y}=\mathrm{7} \\ $$$${x}=\frac{\mathrm{6}}{\mathrm{1}−{y}}=\frac{\mathrm{6}}{\mathrm{1}−\mathrm{7}}=−\mathrm{1} \\ $$$${z}=\:\mathrm{1}+{x}\:=\:\mathrm{1}+\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\therefore\:{x},{y\&z}\:{are}\:−\mathrm{1},\mathrm{7}\:\&\:\mathrm{0}\:{respectively} \\ $$$$ \\ $$$$\partial\epsilon\eta{k}\epsilon\eta\:{la}\int{t}\:\beta\theta{r}\eta \\ $$$$ \\ $$$$ \\ $$

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