Answer
$(x,y) =(6,2)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +&y&=&8\\
x& -&y & =&4
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
1& 1 \\
1& -1
\end{vmatrix}=(1)(-1)-(1)(1)=-1-1=-2$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
8& 1 \\
4& -1
\end{vmatrix}=(8)(-1)-(4)(1)=-8-4=-12$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 8 \\
1& 4
\end{vmatrix}=(1)(4)-(1)(8)=4-8=-4$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\dfrac{-12}{-2}=6$
and
$y=\frac{D_y}{D}=\dfrac{-4}{-2}=2$
Hence, the solution is $(x,y) =(6,2)$.
