Consider the following system of linear equations
$$ \begin{align} x+y+z+w &= 2 \\ x- y+2z+3w &=-10 \\ x -3y+3z+5w &= -7 \\ x+3y- w &= -1 \end{align}$$
If we add the first two (solve for y) or the last two (solve for y), then both are same: $2x+3z+4w=-8$
Does it mean that all the points (a, y, b, c) are the solutions for the system of equations? Here $(a,b,c)$ is the point lying on the plane $2x+3z+4w=-8$ and $y$ is arbitrary.
Whereas, if we compute the row echelon form it says the rank of the coefficients matrix is 2 and the augmented matrix is 3. This implies the solution does not exist.
It seems confusing to me. Any help will be appreciated.
Here is a smaller toy example which, I believe, captures the essence of your problem (it is a system of four equations in two unknowns which is clearly inconsistent):
$$\begin{align} x+y&=1\\ x+y&=2\\ x+y&=1\\ x+y&=2\\ \end{align}$$
Adding either the first or the last pair of equations together yields the line $2x+2y=3$. But you can't conclude anything about the solution space from this single equation. Really all you are saying is that if $(x,y)$ solves a pair of original equations, then it solves the new equation, but the converse is not true at all. What you can do with such a combination of two previous equations is substitute it in the original system (this is one of the elementary row operations), but then you still have to solve the system.