## Precalculus (6th Edition) Blitzer

The values of x and y are, $x=\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\text{ and }y=\left( \frac{{{c}_{2}}{{a}_{1}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$.
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 $x=\left( \frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)\text{ and }y=\left( \frac{{{c}_{2}}{{a}_{1}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \right)$.