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There is a question asking which of the following are systems of linear equations in variables $x_1, x_2, x_3, x_4$, and $x_5$.

And there is a choice, a system containing $$\left\{ \begin{array}{}x_1+x_3+x_5 =8\\ x_2+x_4+x_6 =9\\ x_3+x_5+x_7=4\end{array}\right.$$

Is this choice satisfying the system of linear equations in variables $x_1, x_2, x_3, x_4$, and $x_5$? I found $x_6$ and $x_7$, which are not listed in the question. However, my math teacher once said that $x_1$ or $x_2$ are just symbols representing the variable, which could also be $y$. Therefore, I am not sure whether $x_6, x_7$ are the same. Thank you very much.

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closed as unclear what you're asking by José Carlos Santos, Raffaele, Xander Henderson, Jack D'Aurizio, user223391 Oct 9 '17 at 5:27

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  • $\begingroup$ It's an underdetermined system, without more information, you would find it near impossible to solve. $\endgroup$ – Kevin Oct 5 '17 at 8:11
  • $\begingroup$ Yes. But my professor said it can be linear equation as though it may not be solved. $\endgroup$ – Danny Oct 5 '17 at 18:53
  • $\begingroup$ @Kevin: Not sure why you think it would be so difficult. Where did you get stuck? I solved it immediately using Gaußian elimination and introduction of appropriate parameters.. $\endgroup$ – Mårten W Oct 6 '17 at 12:10
  • $\begingroup$ @MårtenW I didn't get stuck, to me it is a system of seven variables. $\endgroup$ – Kevin Oct 6 '17 at 14:52
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YES!

If you consider $x_6$ and $x_7$ as parameters and not as unknowns, then the system in your question is a perfectly fine system of linear equations in $x_1,\ldots,x_5$.


Gaußian elimination shows that $$ \left\{ \begin{aligned} x_1+x_3+x_5 &= 8 \\ x_2+x_4+x_6 &= 9 \\ x_3+x_5+x_7 &= 4 \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} x_1 &= 4 + x_7 \\ x_2 &= t_2 \\ x_3 &= t_3 \\ x_4 &= 9 - x_ 6 - t_2 \\ x_5 &= 4 - x_7 - t_3 \end{aligned} \right., $$ which gives a parametrisation of the solution space in terms of the parameters $t_2,t_3,x_6,x_7\in\mathbb{R}$.

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  • $\begingroup$ Ok, then please add your working to this answer. I'd be keen to see it. $\endgroup$ – Kevin Oct 6 '17 at 13:12
  • $\begingroup$ @Kevin: Thanks for the downvote. I have added an explicit parametrisation of the solution space. $\endgroup$ – Mårten W Oct 7 '17 at 7:49
  • $\begingroup$ I did not downvote, these rep points are irrelevant to me. I actually was keen to see your working, fwiw, I will +1. $\endgroup$ – Kevin Oct 9 '17 at 6:39
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Hint: No matter how you choose to label the variables, this system of equations has 7 distinct variables, not 5.

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  • $\begingroup$ Thank you very much! Got that^_^ $\endgroup$ – Danny Oct 5 '17 at 18:53
  • $\begingroup$ @gandalf61: How do you know that $x_6$ and $x_7$ are not meant to be parameters? $\endgroup$ – Mårten W Oct 6 '17 at 12:11
  • $\begingroup$ I am applying common sense. Using $x_1$ to $x_5$ to denote variables and $x_6$ and $x_7$ to denote parameters would be bizzare, arbitrary and confusing. If you consider $x_6$ and $x_7$ to be paraneters then how do you know that $x_5$ is not also a parameter - in which case there are actually only 4 variables here ? $\endgroup$ – gandalf61 Oct 6 '17 at 13:45
  • $\begingroup$ @gandalf61: My point is precisely that the question is ill-posed and up to interpretation. $\endgroup$ – Mårten W Oct 7 '17 at 13:48

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