Answer:
<em>The solution of the system is:
</em>
Step-by-step explanation:
The given system of equations is.......

So, the augmented matrix will be: ![\left[\begin{array}{cccc}-1&-3&|&-17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%26-3%26%7C%26-17%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, we will transform the augmented matrix to the reduced row echelon form using row operations.
<u>Row operation 1 :</u> Multiply the 1st row by -1. So..........
![\left[\begin{array}{cccc}1&3&|&17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 2:</u> Add -2 times the 1st row to the 2nd row. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&-12&|&-60\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%26-12%26%7C%26-60%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 3:</u> Multiply the 2nd row by
. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 4:</u> Add -3 times the 2nd row to the 1st row. So........
![\left[\begin{array}{cccc}1&0&|&2\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%7C%262%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, from this reduced row echelon form of the augmented matrix, we can get that
and 
So, the solution of the system is: 
3x=45
45:3=15
x=15
good homework
Given:
line C: y = x + 14
line D: y = 3x + 2
(6,20) or (3,11)
line C: y = 6 + 14 = 20 or y = 3 + 14 = 17
line D: y = 3(6) + 2 = 18 + 2 = 20 or y = 3(3) + 2 = 9 + 2 = 11
(6,20), because both lines pass through this point.