Answer
$(x,y) =(4,-2)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
3x& -&6y&=&24\\
5x& +&4y & =&12
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
3&-6 \\
5& 4
\end{vmatrix}=(3)(4)-(5)(-6)=12+30=42$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
24&-6 \\
12& 4
\end{vmatrix}=(24)(4)-(12)(-6)=96+72=168$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
3&24 \\
5& 12
\end{vmatrix}=(3)(12)-(5)(24)=36-120=-84$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{168}{42}=4$
and
$y=\dfrac{D_y}{D}=\dfrac{-84}{42}=-2$
Hence, the solution set is $(x,y) =(4,-2)$.
