The matrix is not properly formatted.
However, I'm able to rearrange the question as:
![\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C-2%263%265%7C3%5C%5C3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
Operations:


Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.
Answer:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The first operation:

This means that the new second row (R2) is derived by:
Multiplying the first row (R1) by 2; add this to the second row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by 2
![2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]](https://tex.z-dn.net/?f=2%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&5&7|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%265%267%7C1%5Cend%7Barray%7D%5Cright%5D)
The second operation:

This means that the new third row (R3) is derived by:
Multiplying the first row (R1) by -3; add this to the third row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by -3
![-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]](https://tex.z-dn.net/?f=-3%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Hence, the new matrix is:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Answer:
C. f(x) = x - 7 all over 4
Step-by-step explanation:
NB: Let f(x) = y
Exchange X and Y
Make y the subject
f(x) = 4x + 7
y = 4x + 7
x = 4y + 7
x - 7 = 4y
x - 7 all over 4 = 4 ÷ 4
y = x - 7 all over 4
The answer is X = 5 because first you have to take away 2x from that side so u minus 2x from 4x that gives u: 72 + 2x = 82 then you have to minus 72 from both sides and that gives you 2x = 10 once you have that you divide both sides by 2 because you want the X by itself and once you get that X = 5. Hope this helped!
The 9 is in the tenths place so it should be 9x1/10. Meaning it would be 100+3x1+9x1/10+3x1,000