Explain why the solutions of the simultaneous equations ax+by+c=0 dx+ey+f=0 have rational expression in {a,b,c,d,e,f}...

Step 1
given simultaneous equations are,
${\displaystyle ax+by+c=0}$ and
${\displaystyle dx+ey+f=0}$
Step 2
given system can be written as,
${\displaystyle ax+by=-c}$ and
${\displaystyle dx+ey\pm f}$
above system is non homogeneous system of equation,
compaire it with ${\displaystyle ax=b}$
$\left[\begin{array}{cc}a& b\\ d& e\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-c\\ -f\end{array}\right]$
when ${\displaystyle ae=bd}$ then we get a solution in integer.
and if ${\displaystyle ae\ne bd}$ then we have to solve system of simultaneous equation by using row transformation.
$\left[\begin{array}{cc}a& b\\ d& e\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-c\\ -f\end{array}\right]$
$\frac{1}{a}{R}_1\Rightarrow \left[\begin{array}{cc}1& \frac{b}{a}\\ d& e\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-\frac{c}{a}\\ -f\end{array}\right]$
${\displaystyle R_2=R_2-d{R}_1}$ gives,
$\left[\begin{array}{cc}1& \frac{b}{a}\\ 0& e-(\frac{bd}{a})\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-\frac{c}{a}\\ -f+cd\end{array}\right]$
by solving x and y we get rational expressions,
so we have a rational expressions in $\{a,b,c,d,e,f\}$ whenever ${\displaystyle ae\ne bd}$.