A linear system has the row echelon form
\begin{pmatrix}A & B & C & \bigl | &M \\
0 & D & E & \bigl | &N \\
0& 0& F & \bigl | &L
\end{pmatrix} with the unique solution
\begin{pmatrix} {\color{red}X} \\
{\color{blue}Y} \\ Z \end{pmatrix}.
Determine the entries
\color{red} X
=
X
\color{blue} Y
=
Y
Z
=
Z
From the 3rd row in the triangle form we read directly
F Z = L, thus Z = Z.
The second row
D Y + E Z = N
rearranged D Y = N - E Z and with the above
Z = Z substituting gives
{\color{blue}Y = Y}.
Using these two values Z = Z and {\color{blue} Y = Y}
in the 1st row
A X + B Y + C Z = M
rearranged A {\color{red}X} = M - B {\color{blue}Y} + C Z
gives eventually
{\color{red}X = X}.