Answer:
The second choice.
Explanation:
We first write the system of equations as an augmented matrix:
![\left(\begin{array}c 9 & -2 & 5\\ -3 & -4 & -4 \end{array}\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/h984txl6gyz0mos7t6yr2ppfyh3100mxr3.png)
Then we take the determinant
of the left side:
![D=\begin{vmatrix}9 & -2 \\ -3 & -4 \\ \end{vmatrix} =(9)(-4)-(-2)(-3)=-42](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pfmuftfbpmawpuvwxk7ag6v226v1vss6k8.png)
Now the solution of
in the system is
![y=(D_y)/(D)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/36wd3wbeoh4razgk4bk3sfqr6i0374z3fl.png)
where
is the determinant of the matrix formed by replacing
column of the left matrix with elements of the right matrix
:
![D_y=\begin{vmatrix}9 & 5 \\ -3 & -4 \\ \end{vmatrix}=(9)(-4)-(5)(-3)=-21](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xspzt9iydfip5svve0a9cbsk7k9xgdqxpr.png)
therefore,
![y=\frac{\begin{vmatrix}9 & 5 \\ -3 & -4 \\ \end{vmatrix}}{D}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u9k4vel7t8xvn1r6o9j4lsq4erwymvuku8.png)
![\boxed{y=\frac{\begin{vmatrix}9 & 5 \\ -3 & -4 \\ \end{vmatrix}}{-42} =(-21)/(-42) =(1)/(2)}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9hhyhjusteopnj9gs8jau3ly08ophvzvnq.png)
which is the second choice.