Answer:
0
Explanation:
We find the determinant of a matrix by the method below. If we have a matrix:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://img.qammunity.org/2020/formulas/mathematics/high-school/kfph702rerq28gt2m23beyl02du4esve1i.png)
The determinant is
![ad-bc](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mvg66wumrsfoplk2o0magj0bsv31pvg9ee.png)
Now, using cramer's rule, we find x-value by the formula:
![x=(D_x)/(D)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nx1jtqye8ieu7x01cfehvkgmcy33dhqvks.png)
Where D is the determinant of the original problem and
is the determinant of the x-value matrix. How do we get those?
To get original matrix and thus D, we set up the matrix as the coefficients of x and y (s) of both the equations and to get matrix of x-value and thus
, we replace the x values of the matrix with the numbers in the right hand side of the 2 equations. We show this below:
To get D:
![\left[\begin{array}{cc}3&4\\1&-6\end{array}\right] \\D=(3)(-6)-(1)(4)=-18-4=-22](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1qipoyikbrukrum2m5nu8kaw3bcyj02lmt.png)
To get
:
![\left[\begin{array}{cc}12&4\\-18&-6\end{array}\right] \\D_x=(12)(-6)-(-18)(4)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rocm3iiy9h9hr58xm0bbudre3lmk6fx3tx.png)
Putting into the formula, we get:
![x=(D_x)/(D)=(0)/(-22)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/egahyvg8kprrr6qu9jk34taeupz48nzu5c.png)
Thus, the value of x is 0