Finding line of intersection between two planes by solving a system of equations Ok, so the question is to find the line of intersection between two planes, given their equations. 
$x+3y+2z=4$
$x-y-z=4$
I know there's the way of using the vector perpendicular to both normals of the planes as the direction vector of the line of intersection and then finding a specific point on the line. 
But I wanted to find the line of intersection by solving the system- so taking one variable as a parameter and putting the two other variables in terms of that variable. So I just randomly chose x to act as the parameter:
$3y+2z=4-x$   
$-y-z=4-x$
$y=x+4-z$
and then I substituted this y to $3y+2z=4-x$
$3x+12-3z+2z=4-x$
$z=8+4x$
then 
$y=x+4-8-4x$
$y=-4-3x$
and I put x=t and so 
$x=t, y=-4-3t, z=8+4t$
But this doesn't correspond with the answer which is $x=4+t, y=-3t, z=4t$. And I think it's because I took the wrong variable as the parameter..? Is there like a set rule for choosing which variable to be the parameter?
 A: Having 
$$
a_1x+b_1y+c_1z= d_1\\
a_2x+b_2y+c_2z=d_2
$$
or
$$
\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2&b_2 & c_2
\end{array}
\right]
\left[
\begin{array}{c}
x\\
y\\
z
\end{array}
\right]
=
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
$$
Now choosing an invertible $2\times 2$ submatrix in $\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2&b_2 & c_2
\end{array}
\right]$
for instance $\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]$
we rearrange the system as
$$
\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]
\left[
\begin{array}{c}
x\\
z
\end{array}
\right]
=
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
-y 
\left[
\begin{array}{c}
b_1\\
b_2
\end{array}
\right]
$$
and then
$$
\left[
\begin{array}{c}
x\\
z
\end{array}
\right]
=
\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]^{-1}
\left(
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
-t 
\left[
\begin{array}{c}
b_1\\
b_2
\end{array}
\right]
\right)\\
y = t
$$
A: Multiplying your second equation by $4$ and adding to the first we get
$$5x=20+y$$ so $$x=4+\frac{y}{5}$$ we $y=5t$ we get
$$x=4+t$$ and $t$ is a real number.
