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**HallsofIvy** One way to solve that is to set it up as an "augmented matrix" and row-reduce:

$\displaystyle \begin{bmatrix}1 & -1 & -1 &| & 7 \\ 1 & -1 & 2 & | & 2\end{bmatrix}$

$\displaystyle \begin{bmatrix}1 & -1 & -1 &| & 7 \\ 0 & 0 & -1 & | & -5\end{bmatrix}$

$\displaystyle \begin{bmatrix}1 & -1 & 0 & | & 12 \\ 0 & 0 & 1 & | & 5\end{bmatrix}$

which says that $\displaystyle x_1- x_2= 12$, so that $\displaystyle x_1= x_2+ 12$ and $\displaystyle x_3= 5$. That line can be written as the parametric equations $\displaystyle x_1= t+ 12$, $\displaystyle x_2= t$, $\displaystyle x_3= 5$. Taking t= 0 gives (12, 0, 5) and taking t= 1 gives (13, 1, 5). The graph is the straight line through those two points.