Answer
Use Cramer’s rule to solve:
$\begin{align}
& x+y=8 \\
& x-y=-2 \\
\end{align}$
We obtain:
$x=\frac{\left| \begin{matrix}
\underline{8} & \underline{1} \\
\underline{-2} & \underline{-1} \\
\end{matrix} \right|}{\left| \begin{matrix}
\underline{1} & \underline{1} \\
\underline{1} & \underline{-1} \\
\end{matrix} \right|}$
$y=\frac{\left| \begin{matrix}
\underline{1} & \underline{8} \\
\underline{1} & \underline{-2} \\
\end{matrix} \right|}{\left| \begin{matrix}
\underline{1} & \underline{1} \\
\underline{1} & \underline{-1} \\
\end{matrix} \right|}$
Work Step by Step
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:
$x=\frac{\left| \begin{matrix}
-8 & 1 \\
-2 & -1 \\
\end{matrix} \right|}{\left| \begin{matrix}
1 & 1 \\
1 & -1 \\
\end{matrix} \right|}$
$y=\frac{\left| \begin{matrix}
1 & -8 \\
1 & -2 \\
\end{matrix} \right|}{\left| \begin{matrix}
1 & 1 \\
1 & -1 \\
\end{matrix} \right|}$
