Respuesta :
We want to use a matrix to solve
-8x - 8y = -16
6x - 9y = -108
Before solving, simplify the equations as follows:
Divide the first equation by -8 to obtain
x + y = 2 (1)
Divide the second equation by 3 to obtain
2x - 3y = -36 (2)
In matrix form, the equations are
[tex] \begin{bmatrix}1&1\\2&-3\end{bmatrix} \begin{\bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}2 \\ -36 \end{bmatrix}[/tex]
The determinant of the matrix is
D = (1)(-3) - (1)(2) = -5
Use Cramer's Rule.
[tex]x = \frac{1}{D} det(\begin{bmatrix}2&1\\-36&-3 \end{bmatrix} ) = \frac{1}{-5} (-6+36) = -6[/tex]
Similalrly,
[tex]y= \frac{1}{-5} det(\begin{bmatrix} 1&2\\2&-36\end{bmatrix} ) = \frac{1}{-5}(-36-4) =8[/tex]
Answer: (-6, 8) or x = -6, y = 8
-8x - 8y = -16
6x - 9y = -108
Before solving, simplify the equations as follows:
Divide the first equation by -8 to obtain
x + y = 2 (1)
Divide the second equation by 3 to obtain
2x - 3y = -36 (2)
In matrix form, the equations are
[tex] \begin{bmatrix}1&1\\2&-3\end{bmatrix} \begin{\bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}2 \\ -36 \end{bmatrix}[/tex]
The determinant of the matrix is
D = (1)(-3) - (1)(2) = -5
Use Cramer's Rule.
[tex]x = \frac{1}{D} det(\begin{bmatrix}2&1\\-36&-3 \end{bmatrix} ) = \frac{1}{-5} (-6+36) = -6[/tex]
Similalrly,
[tex]y= \frac{1}{-5} det(\begin{bmatrix} 1&2\\2&-36\end{bmatrix} ) = \frac{1}{-5}(-36-4) =8[/tex]
Answer: (-6, 8) or x = -6, y = 8
