i see i did copy the question incorrectly, that seven on the second row should be a negative like you say
@mariomintchev.
so the working/solution will be a bit different , However the method will be the same,
The basic idea is that a matrix is a simpler way of representing an system of equations, performing row operations does not change the solution to the system.
We choose row operations that reduce the complicated system by merging information in different rows,
row echelon form is when we have zeroes below the main diagonal( or bridge) and ones as the first terms in each row. from REF we usually continue to Reduced row echelon form because then the solution can be easily be read off
say \[\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1 \end{array}\right]\left[\begin{array}{c}8\tfrac16\\12\tfrac12\\-5\tfrac12 \end{array}\right]\]
which is kinda shorthand for\[\left[\begin{array}{ccc|c}1&0&0&x_1\\0&1&0&x_2\\0&0&1&x_3 \end{array}\right]=\left[\begin{array}{c}8\tfrac16\\12\tfrac12\\-5\tfrac12 \end{array}\right]\]
which equivalent to\[x_1+0+0=8\tfrac16\]\[0+x_2+0=12\tfrac12\]\[0+0+x_3=-5\tfrac12\] which is the solution to the system