91Ó°ÊÓ

The system of linear equations is$\left\{\begin{array}{l}2x+y=-4\\ 3x-2y=-6\end{array}\right.$

We will take the coefficients of the variables to form the determinant $D$

$D=\left|\begin{array}{cc}2& 1\\ 3& -2\end{array}\right|$

Evaluate it,

$$\begin{gathered} D=\left(2\right)(-2)-\left(1\right)\left(3\right) \\ D=-4-3 \\ D=-7 \end{gathered}$$

We will use the constants in place of the $x$coefficients to find the determinant$D_x$

$D_x=\left|\begin{array}{cc}-4& 1\\ -6& -2\end{array}\right|$

Evaluate it

$$\begin{gathered} D_x=(-4)(-2)-\left(1\right)(-6) \\ {D}_x=8+6 \\ {D}_x=14 \end{gathered}$$

We will use the constants in place of the $y$coefficients to find the determinant$D_y$

$D_y=\left|\begin{array}{cc}2& -4\\ 3& -6\end{array}\right|$

Evaluate it

$$\begin{gathered} D_y=\left(2\right)(-6)-(-4)\left(3\right) \\ {D}_y=-12+12 \\ {D}_y=0 \end{gathered}$$

To find $x$,

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{14}{-7} \\ x=-2 \end{gathered}$$

To find $y$,

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{0}{-7} \\ y=0 \end{gathered}$$

The solution for the system of linear equation is $(-2,0)$.

Substitute $x=-2,y=0$in both the equations,

$$\begin{gathered} 2x+y=-4 \\ 2(-2)+0=-4 \\ -4=-4 \end{gathered}$$

This is true.

$$\begin{gathered} 3x-2y=-6 \\ 3(-2)-2\left(0\right)=-6 \\ -6-0=-6 \\ -6=-6 \end{gathered}$$

This is also true.

Therefore, the solution for the system of linear equation is $(-2,0)$