Question

Solve the following system of equations by the method of cross-multiplication:
$x+ay=b$
$ax-by=c$

Answer 1

The given equations can be written as:

$x+ay-b=0$
$ax-by-c=0$

By method of cross multiplication :

$\Rightarrow \frac{x}{a\times (-c)-(-b)\times (-b)}=\frac{y}{a\times (-b)-1\times (-c)}=\frac{1}{1\times (-b)-a\times a}$

$\Rightarrow \frac{x}{-ac-b^2}=\frac{y}{-ab+c}=\frac{1}{-b-a^2}$

So,

$\Rightarrow \frac{x}{-ac-b^2}=\frac{1}{-b-a^2}$

$\Rightarrow x=\frac{ac+b^2}{a^2+b}$

And,

$\Rightarrow \frac{y}{-ab+c}=\frac{1}{-b-a^2}$

$\Rightarrow y=\frac{ab-c}{a^2+b}$

Hence, the solution is $x=\frac{ac+b^2}{a^2+b}$ and $y=\frac{ab-c}{a^2+b}$