Space/Plane equation/Example intersection/Example

We consider the sets

${}E={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in K^{3}\mid 5x-y+3z=0\right\}}\,$

(from example) and

${}F={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in K^{3}\mid 4x+2y-7z=0\right\}}\,,$

and we are interested in the intersection

${}G:=E\cap F={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in K^{3}\mid 5x-y+3z=0{\text{ and }}4x+2y-7z=0\right\}}\,.$

A point lies in this intersection if and only if it fulfills both conditions, that is, both equations (let us call them ${}I$ and ${}II$). Does there exist a "simpler“ description of this intersection set? A point that fulfills both equations does also fulfill the equation that arises when we add the equations together, or when we multiply the equation with a number ${}s\in K$. Such a linear combination of the equations is, for example,

${}4I-5II=-14y+47z=0\,.$

Therefore,

${}{\begin{aligned}G&={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in K^{3}\mid 5x-y+3z=0{\text{ and }}4x+2y-7z=0\right\}}\\&={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in K^{3}\mid 5x-y+3z=0{\text{ and }}-14y+47z=0\right\}},\end{aligned}}$

since we can reconstruct the original second equation from the new second equation and vice versa. Hence, the conditions are equivalent. The advantage of the second description is that the variable ${}x$ does not occur in the new second equation; it has been eliminated. Therefore, we can resolve with respect to ${}y$, and we obtain

${}y={\frac {47}{14}}z\,.$

For ${}x$, we must have

${}x={\frac {1}{5}}y-{\frac {3}{5}}z={\frac {1}{5}}\cdot {\frac {47}{14}}z-{\frac {3}{5}}z={\frac {47}{70}}z-{\frac {42}{70}}z={\frac {1}{14}}z\,.$

Also, these two resolved equations are together equivalent with the original equations, and, therefore, we have

${}G={\left\{{\begin{pmatrix}{\frac {1}{14}}z\\{\frac {47}{14}}z\\z\end{pmatrix}}\mid z\in K\right\}}\,.$

This description yields a more explicit overview of the set ${}G$.