Write the system of equations as an augmented matrix. Then solve for x and y.
-3x-y=11
-9x-2y=34
Solution =
-3x-y=11 *2 both sides
-9x-2y=34
--> -6x-2y=22
-9x-2y=34 subtract second equ from the first
--> 3x=-12
--> x=-4
Substitute -4 in the first equ
--> -6*-4-2y=22
--> 24-2y=22
--> -2y=22-24
--> y=-2/-2=1
Oh sorry something went wrong
The answer is correct I was confused for a second
To write the system of equations as an augmented matrix, we arrange the coefficients of the variables x and y along with the constants on the right side in a rectangular array.
The given system of equations is:
-3x - y = 11
-9x - 2y = 34
To convert this into an augmented matrix, we write down the coefficients of x and y on the left side, and the constants on the right side. The matrix will have two rows, one for each equation, and three columns, one for each variable and one for the constants.
The augmented matrix is:
|-3 -1 | 11 |
|-9 -2 | 34 |
To solve for x and y using this augmented matrix, we can use row operations to simplify it and bring it to row-echelon form or reduced row-echelon form. There are various methods to solve a system of linear equations, such as Gaussian elimination or Gauss-Jordan elimination. I'll use the Gauss-Jordan elimination method.
1. Apply row operations to the augmented matrix to simplify it:
-3R2 + R1 -> R1 (to eliminate the coefficient of x in the first equation)
-9R1 + R2 -> R2 (to eliminate the coefficient of x in the second equation)
The new augmented matrix will be:
| 1 1/3 | -11/3 |
| 0 1 | -1 |
2. Next, perform row operations to further simplify the matrix:
(-1/3)R2 + R1 -> R1 (to eliminate the coefficient of y in the first equation)
The new augmented matrix will be:
| 1 0 | 4 |
| 0 1 | -1 |
3. The resulting augmented matrix is now in reduced row-echelon form, indicating that x = 4 and y = -1.
Therefore, the solution to the given system of equations is x = 4 and y = -1.