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Question Number 208420 by lepuissantcedricjunior last updated on 15/Jun/24

$$\mathit{r}\mathit{e}\mathit{s}\mathit{o}\mathit{u}\mathit{d}\mathit{r}\mathit{e}\mathit{d}\mathit{a}\mathit{n}\mathit{s}{\mathbb{R}}^3$$$$\{\begin{array}{l}\mathit{x}+\mathit{y}=3\\ \mathit{y}+\mathit{z}=5\end{array}\mathit{x}+\mathit{z}=4$$

Commented by A5T last updated on 15/Jun/24

$$Youshouldlearntosignifythatyouchanged$$$$aquestion.$$$$Thiswasyouroriginalquestion:$$$$\{\begin{array}{l}\mathit{x}+\mathit{y}=3\\ \mathit{y}-\mathit{z}=-1\end{array}\mathit{x}+\mathit{z}=4$$$$\left(\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3\\ 4\\ -1\end{array}\right)$$

Answered by A5T last updated on 15/Jun/24

$$(x,3-x,4-x)infinitelymanysolutions$$

Commented by A5T last updated on 15/Jun/24

$$Evenoneequationcanhaveinfinitelymany$$$$solutions:x+y=1$$

Commented by lepuissantcedricjunior last updated on 15/Jun/24

$$no!!!$$$${\mathit{c}}^{\prime }est\mathit{t}\mathit{r}\mathit{o}\mathit{i}\mathit{s}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}$$

Answered by Frix last updated on 15/Jun/24

$$1+2=3$$$$1+3=4$$$$2+3=5$$$$x=1y=2z=3$$

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