How do I choose a free variable? I have a question regarding the Gaussian method for solving linear equations. I had to solve 2 equations with 3 unknowns and naturally with the elimination process I had 2 variables left. I thought that it didn't matter which variable I chose as a free variable, but apparently after inputting random values after choosing the last variable as the free one, the values didn't match. So is there any method of how to choose a free variable? 
\begin{cases}
 & 4x_{1} + 2x_{2} + 3x_{3} = -2 \\
 & 2x_{1} + 8x_{2} - x_{3} = -10
\end{cases}
\begin{pmatrix}
    4 & 2 & 3 & | & -2 \\
    2 & 8 & -1 & | & -10
\end{pmatrix}
\begin{pmatrix}
    2 & 8 & -1 & | & -10 \\
    4 & 2 & 3 & | & -2
\end{pmatrix}
\begin{pmatrix}
    2 & 8 & -1 & | & -10 \\
    0 & -14 & 5 & | & 18
\end{pmatrix}
$$
-14x_{2} + 5x_{3} = 18
$$
$$
x_{3} = C
$$
$$
x_2 = \frac{18-5C}{-14}
$$
$$
4x_1 = -2 -3C - 2*\frac{18-5C}{-14}
$$
$$
4x_1 = -2 -3C - \frac{18-5C}{-7}
$$
$$
x_1 = \frac{-14-21C-18+5C}{28}
$$
$$
x_1 = \frac{-16C-32}{28} = \frac{-8C-16}{14}
$$
this is what I did, but using x3 as a free variable was not a correct choice, because after inputting a random value, the first equation didn't equal to -2. so I want to find out why do I necessarily have to choose x2 instead of x3.

As noted by Bye_World in the comments, in Reduced Row Echelon Form, the matrix specified becomes:
\begin{pmatrix}
    1 & 0 & \frac{13}{14} & | & \frac{1}{7} \\
    0 & 1 & -\frac{5}{14} & | & -\frac{9}{7}
\end{pmatrix}
You then note that you can use $x_3$ as the free variable. So you get:
$$x_1 = -\frac{13}{14}x_3 + \frac{1}{7}$$
$$x_2 = \frac{5}{14}x_3 + -\frac{9}{7}$$
Hope this helps!
