Solve system of equations using Cramer’s Rule.

$\left\{\begin{array}{l}3x+8y=-3\\ 2x+5y=-3\end{array}\right.$

The system of linear equations is$\left\{\begin{array}{l}3x+8y=-3\\ 2x+5y=-3\end{array}\right.$

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

$D=\left|\begin{array}{cc}3& 8\\ 2& 5\end{array}\right|$

Evaluate it,

$$\begin{gathered} D=\left(3\right)\left(5\right)-\left(2\right)\left(8\right) \\ D=15-16 \\ D=-1 \end{gathered}$$

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

$D_x=\left|\begin{array}{cc}-3& 8\\ -3& 5\end{array}\right|$

Evaluate it

$$\begin{gathered} D_x=(-3)\left(5\right)-(-3)\left(8\right) \\ {D}_x=-15+24 \\ {D}_x=9 \end{gathered}$$

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

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

Evaluate it

$$\begin{gathered} D_y=\left(3\right)(-3)-\left(2\right)(-3) \\ {D}_y=-9+6 \\ {D}_y=-3 \end{gathered}$$

To find $x$,

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{9}{-1} \\ x=-9 \end{gathered}$$

To find $y$,

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{-3}{-1} \\ y=3 \end{gathered}$$

The solution for the system of linear equation is$(-9,3)$

Substitute $x=-9,y=3$in both the equations,

$$\begin{gathered} 3x+8y=-3 \\ 3(-9)+8\left(3\right)=-3 \\ -27+24=-3 \\ -3=-3 \end{gathered}$$

This is true.

$$\begin{gathered} 2x+5y=-3 \\ 2(-9)+5\left(3\right)=-3 \\ -18+15=-3 \\ -3=-3 \end{gathered}$$

This is also true.

Therefore, the solution for the system of linear equation is $(-9,3)$