Below is a problem I did. I believe my answer is right but I am not sure. Also, is my notation right?
Problem:
Solve the following system of equations:
\begin{align*}
w + x + y - z &= 1 \\
2w + x + y + z &= 8 \\
2w + x + y - 3z &= -4 \\
w + x + y + z &= 7
\end{align*}
Answer:
To solve these equations, I setup a matrix with a slight change in the order of the equations to make the computation easier. $$ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} $$ Now we apply row operations to simplify the matrix. \begin{align*} \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & -2 & -6 \\ 0 & -1 & -1 & -1 & -6 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & -1 & -1 & -1 & -6 \\ 0 & -1 & -1 & -5 & -18 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & -1 & -1 & -1 & -6 \\ 0 & -1 & -1 & -5 & -18 \\ 0 & 0 & 0 & 1 & 3 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & -1 & -1 & 0 & -3 \\ 0 & -1 & -1 & -5 & -18 \\ 0 & 0 & 0 & 1 & 3 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & -1 & -1 & 0 & -3 \\ 0 & -1 & -1 & 0 & -3 \\ 0 & 0 & 0 & 1 & 3 \\ \end{bmatrix} \\ \end{align*} At this point we know that $z = 3$ and that we are not going to have a unique solution. \begin{align*} \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 0 & -1 & -1 & 0 & -3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 3 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 0 & 4 \\ 0 & -1 & -1 & 0 & -3 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} \\ \begin{bmatrix} 1 & 1 & 1 & 1 & 7 \\ 1 & 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 1 & 8 \\ 2 & 1 & 1 & -3 & -4 \\ \end{bmatrix} &\sim \begin{bmatrix} 1 & 1 & 1 & 0 & 4 \\ 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} \\ \end{align*} Now we have found $w = 1$. Now we work with the equation represented by the first row of the matrix. \begin{align*} w + x + y &= 4 \\ 1 + x + y &= 4 \\ x + y &= 3 \\ x &= 3 - y\\ \end{align*} Let $y_0$ be any arbitrary real number. Then the solution set for this system of equations is: $$ ( 1, 3 - y_0, y_0, 3 ) $$
Your process is no doubt correct. But you can try this.
Solution:
$x+y=:v$
Now system of equations reduces:
$$w + v - z = 1 \\ 2w + v + z = 8 \\ 2w + v - 3z = -4 \\ w + v + z = 7$$
We can represent the system of linear equations in matrix form follows (without taking third equation):
$$ \begin{bmatrix} 1 & 1 & -1 \\ 2 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} w \\ v \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ 7 \\ \end{bmatrix} $$
Or,
$$ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 1 & 1 & 8 \\ 1 & 1 & 1 & 7 \\ \end{array}\right] $$
Step 1: Subtract twice the first row from the second row and once the first row from the third row.
$$ \left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & -1 & 3 & 6 \\ 0 & 0 & 2 & 6 \\ \end{array}\right]\sim\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & -1 & 3 & 6 \\ 0 & 0 & 1 & 3 \\ \end{array}\right] $$
Step 2: Add the second row to the first row.
$$ \left[\begin{array}{ccc|c} 1 & 0 & 2 & 7 \\ 0 & -1 & 3 & 6 \\ 0 & 0 & 1 & 3 \\ \end{array}\right] $$
Step 3: Subtract three times the third row from the second row and subtract twice third row from the first row.
$$ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -3 \\ 0 & 0 & 1 & 3 \\ \end{array}\right] $$
Now, we can read the solutions for the variables:
$$w = 1\\ v = 3\\ z = 3\\$$
Check solution for third equation: $2w + v - 3z = 2+3-9=-4$ satisfy the third equation.
Now, $v=x+y=3$. Hence, there is infinitely many solution for $x,y$.