Question

Solve for $x$ and $y$
$ax+by=c$
$bx+ay=1+c$

Answer 1

$ax+by=c\Rightarrow ax+by-c=0bx+ay=1+c\Rightarrow bx+ay-(1+c)=0$

On cross multiplication$,$

$\frac{x}{\left|\begin{array}{cc}b& -c\\ a& -(1+c)\end{array}\right|}=\frac{-y}{\left|\begin{array}{cc}a& -c\\ b& -(1+c)\end{array}\right|}=\frac{1}{\left|\begin{array}{cc}a& b\\ b& a\end{array}\right|}\Rightarrow \frac{x}{-b(1+c)+ac}=\frac{-y}{-a(1+c)+cb}=\frac{1}{a^2-b^2}\therefore x=\frac{ac-b-bc}{a^2-b^2}y=\frac{a+ac-cb}{a^2-b^2}$