Answer
$(x,y) =\left(-\frac{13}{4},2\right)$
Work Step by Step
Use zero for the missing variable.
The given system of equations is
$\left\{\begin{matrix}
4 x& +&5y&=&-3\\
0x& -&2y & =&-4
\end{matrix}\right.$
Determinant $D$ consists of the $x$ and $y$ coefficients.
$D=\begin{vmatrix}
4&5 \\
0& -2
\end{vmatrix}=(4)(-2)-(0)(5)=-8-0=-8$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-3&5 \\
-4& -2
\end{vmatrix}=(-3)(-2)-(-4)(5)=6+20=26$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
4&-3 \\
0& -4
\end{vmatrix}=(4)(-4)-(0)(-3)=-16+0=-16$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{26}{-8}=-\dfrac{13}{4}$
and
$y=\dfrac{D_y}{D}=\dfrac{-16}{-8}=-2$
Hence, the solution set is $(x,y) =\left(-\frac{13}{4},2\right)$.
