## Precalculus (6th Edition) Blitzer

Use Cramer’s rule to solve: \begin{align} & x+y=8 \\ & x-y=-2 \\ \end{align} We obtain: $x=\frac{\left| \begin{matrix} \underline{8} & \underline{1} \\ \underline{-2} & \underline{-1} \\ \end{matrix} \right|}{\left| \begin{matrix} \underline{1} & \underline{1} \\ \underline{1} & \underline{-1} \\ \end{matrix} \right|}$ $y=\frac{\left| \begin{matrix} \underline{1} & \underline{8} \\ \underline{1} & \underline{-2} \\ \end{matrix} \right|}{\left| \begin{matrix} \underline{1} & \underline{1} \\ \underline{1} & \underline{-1} \\ \end{matrix} \right|}$
Cramer rule states that for any linear equations: \begin{align} & {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ & {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \end{align} The value for $x,y$ will be calculated as here: $x=\frac{\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\ {{c}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}$ $y=\frac{\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|}$ Therefore, the values of $x,y$ for the provided linear equations are: $x=\frac{\left| \begin{matrix} -8 & 1 \\ -2 & -1 \\ \end{matrix} \right|}{\left| \begin{matrix} 1 & 1 \\ 1 & -1 \\ \end{matrix} \right|}$ $y=\frac{\left| \begin{matrix} 1 & -8 \\ 1 & -2 \\ \end{matrix} \right|}{\left| \begin{matrix} 1 & 1 \\ 1 & -1 \\ \end{matrix} \right|}$