\begin{cases} x+ 2y - z + w - t = 0 \\ x - y + z + 3w - 2t = 0 \end{cases}
Add 1st to the 2nd:
$$2x + y + 4w - 3t = 0 \\ y = -2x - 4w + 3t = 0$$
Substitute y in the 1st:
$$x - 4x - 8w + 6t - z + w - t = 0 \\ z = -3x - 7w + 5t$$
Both z and y in terms of x,w,t. Writing using matrix form:
$$\left[\begin{matrix} x \\ y \\ z \\ w \\ t \\ \end{matrix}\right] = x\left[\begin{matrix} 1 \\ -2 \\ -3 \\ 0 \\ 0 \\ \end{matrix}\right] + w\left[\begin{matrix} 0 \\ -4 \\ -7 \\ 1 \\ 0 \\ \end{matrix}\right] + t\left[\begin{matrix} 0 \\ 3 \\ 5 \\ 0 \\ 1 \\ \end{matrix}\right] $$
Solution $x(1,-2,-3,0,0) + w(0,-4,-7,1,0) + t(0,3,5,0,1)$
The previous question was to check the correctness of that solution. I've multiplied each vector by the matrix formed by the coefficients of the linear system and the result is 0 for all. But now I'm seeing that the choice of isolating y and z was arbitrary, I could have isolated any pair of unknowns. Are all linear combinations of three vectors found this way part of the same set of solutions?