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$$\begin{align*}x_1 + x_2 &= a\\y_1 + y_2 &= b\\z_1 + z_2 &= c\\x_1 + y_1 + z_1 &= 0\\x_2 &= z_2\end{align*}$$

Solve for $x_1, x_2, y_1, y_2, z_1, z_2$ in terms of $a, b$ and $c$.

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    $\begingroup$ You will still have one free parameter because you have one too few equations. The usual substitution technique will get you there. What have you tried? Where are you stuck? $\endgroup$ – Ross Millikan Nov 28 '17 at 15:40
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Construct it as a matrix and just Gauss it down.

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  • $\begingroup$ I tried that, but the matrix being 5x6 (not square) makes it a lot more difficult $\endgroup$ – Koy Nov 28 '17 at 15:39
  • $\begingroup$ @Koy Yes this is something you will have to live with. You have 6 unknowns and only 5 equations, it is unsolvable (you always need the same number of equations as the number of variables you are trying to solve for). In other words, you will have variables which you can only express in terms of other variables, a dependency on one another. The solution is not a point, it is a line or a space. To solve it exactly you need to guess a solution or use numerical methods. $\endgroup$ – digestivee Nov 30 '17 at 13:22
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You can either solve it by brute force, substituting between the simpler equations first which will eventually lead you to a solution.

An alternative method, if you are familiar with Linear Algebra courses, is expressing the system via matrices in the form $Ax=b$, where $A$ is the matrix of coefficients of the variables, $x$ the $1\times n$ matrix of $n$ variables and $b$ the $1\times n$ right hand matrix of the equations. After that, you can apply the Gauss Elimination Method to yield your solution.

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  • $\begingroup$ Could you demonstrate the solution using the matrix? Because I've tried that but I end up circling and getting the same rows over and over again, just as with the brute force solution I keep getting the same equations. $\endgroup$ – Koy Nov 28 '17 at 16:00
  • $\begingroup$ @Koy Hi ! As you can observe from the last equation of your system, $x_2 = z_2$, you will have a free parameter because there are one few too much equations. That should not make you confused though, you can proceed with the solutions and let the extra parameter be a parameter ! $\endgroup$ – Rebellos Nov 28 '17 at 20:51

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