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Solve the system

$$x_1+x_2+2x_3=3\\ 6x_1+7x_2-3x_3=-3$$


$$\begin{bmatrix} 1 & 1 & 2 & 3\\ 6 & 7 & -3 & -3\\ \end{bmatrix}\text{~}\begin{bmatrix} 1 & 1 & 2 & 3\\ 0 & 1 & -15 & -21\\ \end{bmatrix}\text{~}\begin{bmatrix} 1 & 0 & 17 & 24\\ 0 & 1 & -15 & -21\\ \end{bmatrix}$$

And using this I found that: $$x_1+17x_3=24\\ x_2-15x_3=-21$$

As far as I know this is correct, but my issue is how to state my solution.

$$\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} ?\\ ?\\ ? \end{bmatrix}+\begin{bmatrix} ?\\ ?\\ ? \end{bmatrix}s$$

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Just like we did here. – JohnD Jan 27 '13 at 19:44
up vote 1 down vote accepted

Notice that all of your equations have a common factor, namely $x_3$.

Rewrite $x_1$ as a function of $x_3$ and then rewrite $x_2$ as a function of $x_3$. Now by letting $x_3$ vary over your the given field, you'll attain all solutions.

You'll get $x_1=24-17x_3$ and $x_2=-21+15x_3$, so your set of solutions should be $\{ \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T : x_1, x_2, x_3\in F \wedge x_1=24-17x_3 \wedge x_2=-21+15x_3\}$ or more succinctly $\{ \begin{bmatrix} 24-17x_3 \space -21+15x_3 \space x_3 \end{bmatrix}^T : x_3\in F\}$, where $F$ is the field you're working on.

In this answer I assumed that your calculations are correct.

For similar problems check this and this.

EDIT: Your calculations are correct.

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And then to put it in the form OP wants, $$\pmatrix{24-17x_3\cr-21+15x_3\cr x_3\cr}=\pmatrix{24\cr-21\cr0}+\pmatrix{-17\cr15\cr1\cr}x_3$$ – Gerry Myerson Jan 28 '13 at 5:40
    
yes, thank you. – ground.clouds1 Feb 11 '13 at 1:47

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