 UkusakazaL

2021-09-22

Explain why the solutions of the simultaneous equations

have rational expression in $\left\{a,b,c,d,e,f\right\}$ whenever $ae\ne bd$. Bertha Stark

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

Do you have a similar question?