Answer
$(\displaystyle \frac{c_{1}b_{2}-c_{2}b_{1}}{a_{1}b_{2}-a_{2}b_{1}},\ \ \frac{c_{1}a_{2}-c_{2}a_{1}}{b_{1}a_{2}-b_{2}a_{1}}),\quad a_{1}b_{2}\neq a_{2}b_{1}$
Work Step by Step
Eliminating x,
multiply the first equation with $a_{2}$ and the second with $-a_{1}$
$\left\{\begin{array}{l}
a_{1}a_{2}x+b_{1}a_{2}y=c_{1}a_{2}\\
-a_{1}a_{2}x-b_{2}a_{1}y=-c_{2}a_{1}
\end{array}\right.$
add the equations
$(b_{1}a_{2}-b_{2}a_{1})y=c_{1}a_{2}-c_{2}a_{1}$
$y=\displaystyle \frac{c_{1}a_{2}-c_{2}a_{1}}{b_{1}a_{2}-b_{2}a_{1}},\quad a_{1}b_{2}\neq a_{2}b_{1}$
Similarly, eliminating y,
multiply the first equation with $b_{2}$ and the second with $-b_{1}$
$\left\{\begin{array}{l}
a_{1}b_{2}x+b_{1}b_{2}y=c_{1}b_{2}\\
-a_{2}b_{1}x-b_{2}b_{1}y=-c_{2}b_{1}
\end{array}\right.$
add the equations
$(a_{1}b_{2}-a_{2}b_{1})x=c_{1}b_{2}-c_{2}b_{1}$
$x=\displaystyle \frac{c_{1}b_{2}-c_{2}b_{1}}{a_{1}b_{2}-a_{2}b_{1}},\quad a_{1}b_{2}\neq a_{2}b_{1}$
