Given the following System of Equations:
[tex]\begin{cases}x+2y=8 \\ -3x-2y=12\end{cases}[/tex]
You can solve it with Cramer's Rule. The steps are shown below:
1. By definition, you know that for "x"
[tex]x=\frac{D}{D_x}=\frac{\begin{bmatrix}{c_1} & {b_1} & {} \\ {c_2_{}_{}_{}} & {b_2} & {} \\ {} & {} & \end{bmatrix}}{\begin{bmatrix}{a1_{}} & {b_1} & {} \\ {a_2_{}} & {b_2} & {} \\ {} & {} & \end{bmatrix}}[/tex]
In this case:
[tex]\begin{gathered} c_1=8 \\ c_2=12_{} \\ b_1=2 \\ b_2=-2_{} \\ a_1=1 \\ a_2=-3 \end{gathered}[/tex]
Then, you can substitute values and evaluating, you get that the value of "x" is:
[tex]x=\frac{\begin{bmatrix}{8_{}} & {2_{}} & {} \\ {12_{}} & {-2_{}} & {} \\ {} & {} & \end{bmatrix}}{\begin{bmatrix}{1_{}} & {2_{}} & {} \\ {-3_{}} & {-2_{}} & {} \\ {} & {} & \end{bmatrix}}=\frac{(-2)(8)-(2)(12)}{(-2)(1)-(2)(-3)}=\frac{-16-24}{-2+6}=-10[/tex]
2. By definition, for "y":
[tex]y=\frac{D_y}{D}=\frac{\begin{bmatrix}{a_1} & {c_1} & {} \\ {a_2} & {c_2} & {} \\ {} & {} & {}\end{bmatrix}}{\begin{bmatrix}{a_1} & {b_1} & {} \\ {a_2} & {b_2} & {} \\ {} & {} & {}\end{bmatrix}}[/tex]
Knowing the values, substitute and evaluate:
[tex]y=\frac{\begin{bmatrix}{1_{}} & {8_{}} & {} \\ {-3_{}} & {12_{}} & {} \\ {} & {} & {}\end{bmatrix}}{\begin{bmatrix}{1_{}} & {2_{}} & {} \\ {-3_{}} & {-2_{}} & {} \\ {} & {} & {}\end{bmatrix}}=\frac{(12)(1)-(-3)(8)}{(-2)(1)-(-3)(2)}=\frac{12+24}{-2+6}=9[/tex]
Therefore, the answer is:
[tex]\begin{gathered} x=\frac{\begin{bmatrix}{8_{}} & {2_{}} & {} \\ {12_{}} & {-2_{}} & {} \\ {} & {} & \end{bmatrix}}{\begin{bmatrix}{1_{}} & {2_{}} & {} \\ {-3_{}} & {-2_{}} & {} \\ {} & {} & \end{bmatrix}}=\frac{-16-24}{-2+6}=-10 \\ \\ \\ y=\frac{\begin{bmatrix}{1_{}} & {8_{}} & {} \\ {-3_{}} & {12_{}} & {} \\ {} & {} & {}\end{bmatrix}}{\begin{bmatrix}{1_{}} & {2_{}} & {} \\ {-3_{}} & {-2_{}} & {} \\ {} & {} & {}\end{bmatrix}}=\frac{12+24}{-2+6}=9 \end{gathered}[/tex]
