Solve system of equations using Cramer’s Rule.

$\left\{\begin{array}{l}x-3y=-9\\ 2x+5y=4\end{array}\right.$

The system of linear equations is$\left\{\begin{array}{l}x-3y=-9\\ 2x+5y=4\end{array}\right.$

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

$D=\left|\begin{array}{cc}1& -3\\ 2& 5\end{array}\right|$

Evaluate it,

$$\begin{gathered} D=\left(1\right)\left(5\right)-(-3)\left(2\right) \\ D=5+6 \\ D=11 \end{gathered}$$

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

$D_x=\left|\begin{array}{cc}-9& -3\\ 4& 5\end{array}\right|$

Evaluate it

$$\begin{gathered} D_x=(-9)\left(5\right)-(-3)\left(4\right) \\ {D}_x=-45+12 \\ {D}_x=-33 \end{gathered}$$

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

$D_y=\left|\begin{array}{cc}1& -9\\ 2& 4\end{array}\right|$

Evaluate it

$$\begin{gathered} D_y=\left(1\right)\left(4\right)-(-9)\left(2\right) \\ {D}_y=4+18 \\ {D}_y=22 \end{gathered}$$

To find $x$,

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-33}{11} \\ x=-3 \end{gathered}$$

To find $y$,

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{22}{11} \\ y=2 \end{gathered}$$

The solution for the system of linear equation is$(-3,2)$

Substitute $x=-3.y=2$in both the equations,

$$\begin{gathered} x-3y=-9 \\ -3-3\left(2\right)=-9 \\ -3-6=-9 \\ -9=-9 \end{gathered}$$

This is true.

$$\begin{gathered} 2x+5y=4 \\ 2(-3)+5\left(2\right)=4 \\ -6+10=4 \\ 4=4 \end{gathered}$$

This is also true.

Therefore, the solution for the system of linear equation is $(-3,2)$