Answer:
Solution : (15, - 11)
Step-by-step explanation:
We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )
![\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D-4%26-5%26%7C%26-5%5C%5C%20-6%26-8%26%7C%26-2%5Cend%7Bbmatrix%7D)
Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )
Row Echelon Form :
![\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D1%5C%3A%26%5C%3A%5Ccdots%20%5C%3A%26%5C%3Ab%5C%3A%5C%5C%200%5C%3A%26%5Cddots%20%5C%3A%26%5C%3A%5Cvdots%20%5C%5C%200%5C%3A%26%5C%3A0%5C%3A%26%5C%3A1%5Cend%7Bpmatrix%7D)
Step # 1 : Swap the first and second matrix rows,
![\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D-6%26-8%26-2%5C%5C%20-4%26-5%26-5%5Cend%7Bpmatrix%7D)
Step # 2 : Cancel leading coefficient in row 2 through
,
![\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bpmatrix%7D-6%26-8%26-2%5C%5C%200%26%5Cfrac%7B1%7D%7B3%7D%26-%5Cfrac%7B11%7D%7B3%7D%5Cend%7Bpmatrix%7D)
Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.
![\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%260%26%7C%2615%5C%5C%200%261%26%7C%26-11%5Cend%7Bbmatrix%7D)
As you can see our solution is x = 15, y = - 11 or (15, - 11).