Answer
The values of x and y are,
$x=\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\text{ and }y=\left( \frac{{{c}_{2}}{{a}_{1}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$.
Work Step by Step
From equation (II), $y=\frac{{{e}_{2}}-{{a}_{2}}x}{{{b}_{2}}}$; we place it into equation (I),
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}\left( \frac{{{c}_{2}}-{{a}_{2}}x}{{{b}_{2}}} \right)={{c}_{1}} \\
& {{a}_{1}}{{b}_{2}}x+{{b}_{1}}{{c}_{2}}-{{a}_{2}}{{b}_{1}}x={{c}_{1}}{{b}_{2}} \\
& ({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})x={{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}} \\
& x=\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)
\end{align}$
Subtitute this value of x in equation (I),
${{a}_{1}}\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)+{{b}_{1}}y={{c}_{1}}$
Solve to get,
$y=\left( \frac{{{c}_{2}}{{a}_{1}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$
Therefore, the solution of the given linear system is $x=\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\text{ and }y=\left( \frac{{{c}_{2}}{{a}_{1}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$.
