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

$$$$$$\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n};$$$$\begin{array}{c}\mathit{x}+\mathit{y}-\mathit{z}=5\\ \mathit{z}-\mathit{y}\mathit{x}=7\\ \mathit{z}=1+\mathit{x}\end{array}\}\mathit{x};\mathit{y};\mathit{z}=??$$$$$$

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

$$\begin{array}{c}\mathit{x}+\mathit{y}-\mathit{z}=5...\left(i\right)\\ \mathit{z}-\mathit{y}\mathit{x}=7...\left(ii\right)\\ \mathit{z}=1+\mathit{x}....\left(iii\right)\end{array}\}\mathit{x};\mathit{y};\mathit{z}=??$$$$\left(i\right)\Rightarrow x+y-(1+x)=5$$$$\Rightarrow y=6$$$$\left(ii\right)\Rightarrow z-6x=7$$$$\left(iii\right)\Rightarrow z-x=1$$$$-5x=6\Rightarrow x=-6/5$$$$z=1+x=1-\frac{6}{5}=-\frac{1}{5}$$$$(x,y,z)=(-\frac{6}{5},6,-\frac{1}{5})$$

Answered by Yhusuph last updated on 03/Mar/23

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

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