## Precalculus (6th Edition) Blitzer

The ${{D}_{x}}\text{ and }{{D}_{y}}$ are represented in terms of the coefficients and constants.
Given a linear system in two variables, \begin{align} & {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ & {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \\ \end{align} Where $a_1$, $a_2$, $b_1$, $b_2$ are coefficients and $c_1$ and $c_2$ are constants. Then, $x=\frac{\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}$ And $y=\frac{\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}\text{ }$ Where, $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\ne 0$ Or $x=\frac{{{D}_{x}}}{D}\text{ and }y=\frac{{{D}_{y}}}{D}$ Where, \begin{align} & {{D}_{x}}=\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right| \\ & \text{ and } \\ & {{D}_{{}}}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\ne 0 \\ \end{align} So, ${{D}_{x}}\text{ and }{{D}_{y}}$ are represented in terms of the coefficients and constants.