We have the augmented matrix of a system of equations:
$$ \begin{pmatrix} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -7 \end{pmatrix} $$
The system has infinitely many solutions with solutions as $(x,y,z) = (s,-7,4-s)$
and another system:
$$ \begin{pmatrix} 1 & 0 & 0& 0 &2\\ 0 & 1 & 0 & 0&-5\\ 0 & 0 & 1 &1 &7\\ 0 & 0 & 0 & 0 &0 \end{pmatrix} $$
The system has infinitely many solutions with solutions as $$(x,y,z,p) = (2,-5,s,7-s)$$
Is the two solutions correct?
There is a slightly different approach as $(x,y,z)$ could calculate as $(4,-7,0)$. But then any value can be assigned to z if the remaining values are re-calculated in the row reduction. The value y is re-calculated first while the value $x$ is re-calculated last.
Then $(x,y,z,p)$ could calculate as $(2,-5,7,0)$. However, any value can be assigned to p if the remaining values are re-calculated in the row reduction. The value $z$ is re-calculated first, the value $y$ is re-calculated second and the value $x$ is re-calculated last.
Well, begin with the augmented matrix of
$\begin{pmatrix} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -7 \\ 0 & 1 & 0 & -7 \end{pmatrix}$
And the row reduction is
$\begin{pmatrix} 1 & 0 & 1 & 4 \\ 0 & 1 & 0 & -7 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
Then since the value of $z$ is to be determined on the third row of the row reduction, an indication of $0 = 0$ on the third row means that $z = 0$. Or $z$ = any value because $0 * z = 0$ is a valid operation.