(related to Definition: Linear Equations with many Unknowns)
Let $n\ge 1$ and let $\alpha_1x_1+\ldots+\alpha_nx_n=\gamma$ be a linear equation with the unknowns $x_1,\ldots,x_n$ and the elements $\alpha_1,\ldots,\alpha_n,\gamma$ of a field $F.$ There are the following cases for finding a solution to this equation:
We can re-index the coefficients and unknowns appropriately to let $\alpha_1=0,\ldots,\alpha_{m-1}=0$ and $\alpha_{m}\neq 0,\alpha_{m+1}\neq 0,\ldots,\alpha_n\neq 0.$ The (homogenous or inhomogenous) equation is then given by $$0 x_1 + 0 x_{m-1} + \alpha_m x_m + \alpha_{m+1} x_{m+1} + \ldots +\alpha_n x_n=\gamma.$$ Then $$x_m:=\frac{\gamma-\alpha_{m+1}x_{m+1}-\ldots-\alpha_n x_n}{\alpha_m}\quad ( * )$$ is a solution for any choice of the $n-1$ unknowns $x_1,\ldots,x_{m-1},x_{m+1}\ldots,x_n\in F.$ We say that the solution $( * )$ has $n-1$ degrees of freedom. If $\gamma=0$ (the equation is homogenous), then we call the special choice $x_1=0,\ldots,x_n=0$ the trivial solution of $( * ).$
The inhomogenous equation is then given by $$0 x_1 + \ldots +0x_n=\gamma.$$ In this case, for any choice of the $n$ unknowns $x_1,\ldots,x_n\in F,$ the equation has no solution.
The homogenous equation is then given by $$0 x_1 + \ldots +0x_n=0$$
Any choice of the $n$ unknowns $x_1,\ldots,x_n\in F,$ is a solution of the equation.
Proofs: 1
Examples: 1