quality:correction
An example for the equation version

Suppose given n {\displaystyle n} equations and a starting point x 0 {\displaystyle \mathbf {x} _{0}}.At any step in a Gauss-Seidel iteration, solve the first equation for x 1 {\displaystyle x_{1}} in terms of x 2, …, x n {\displaystyle x_{2},\dots,x_{n}}; then solve the second equation for x 2 {\displaystyle x_{2}} in terms of x 1 {\displaystyle x_{1}} just found and the remaining x 3, …, x n {\displaystyle x_{3},\dots,x_{n}}; and continue to x n {\displaystyle x_{n}}. Then, repeat iterations until (hopefully) converged.

To make it clear, consider an example: 10 x 1 − x 2 + 2 x 3 = 6, − x 1 + 11 x 2 − x 3 + 3 x 4 = 25, 2 x 1 − x 2 + 10 x 3 − x 4 = − 11, 3 x 2 − x 3 + 8 x 4 = 15. {\displaystyle {\begin{array}{rrrrl}10x_{1}&-x_{2}&+2x_{3}&&=6,\\-x_{1}&+11x_{2}&-x_{3}&+3x_{4}&=25,\\2x_{1}&-x_{2}&+10x_{3}&-x_{4}&=-11,\\&3x_{2}&-x_{3}&+8x_{4}&=15.\end{array}}}

Solving for x 1, x 2, x 3 {\displaystyle x_{1},x_{2},x_{3}} and x 4 {\displaystyle x_{4}} gives: x 1 = x 2 / 10 − x 3 / 5 + 3 / 5, x 2 = x 1 / 11 + x 3 / 11 − 3 x 4 / 11 + 25 / 11, x 3 = − x 1 / 5 + x 2 / 10 + x 4 / 10 − 11 / 10, x 4 = − 3 x 2 / 8 + x 3 / 8 + 15 / 8. {\displaystyle {\begin{aligned}x_{1}&=x_{2}/10-x_{3}/5+3/5,\\x_{2}&=x_{1}/11+x_{3}/11-3x_{4}/11+25/11,\\x_{3}&=-x_{1}/5+x_{2}/10+x_{4}/10-11/10,\\x_{4}&=-3x_{2}/8+x_{3}/8+15/8.\end{aligned}}}

Suppose we choose (0, 0, 0, 0) as the initial approximation, then the first approximate solution is given by x 1 = 3 / 5 = 0.6, x 2 = (3 / 5) / 11 + 25 / 11 = 3 / 55 + 25 / 11 = 2.3272, x 3 = − (3 / 5) / 5 + (2.3272) / 10 − 11 / 10 = − 3 / 25 + 0.23272 − 1.1 = − 0.9873, x 4 = − 3 (2.3272) / 8 + (− 0.9873) / 8 + 15 / 8 = 0.8789. {\displaystyle {\begin{aligned}x_{1}&=3/5=0.6,\\x_{2}&=(3/5)/11+25/11=3/55+25/11=2.3272,\\x_{3}&=-(3/5)/5+(2.3272)/10-11/10=-3/25+0.23272-1.1=-0.9873,\\x_{4}&=-3(2.3272)/8+(-0.9873)/8+15/8=0.8789.\end{aligned}}}

Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after four iterations.

The exact solution of the system is (1, 2, −1, 1).