when you have $n$ equations and $m$ variables that $n<m$ you must consider $m-n$ variables free and fixed (they can take arbitrary value)
then you try to solve your equation with $n$ equations and $n$ variables
as at this problem you can do it:
$w+x+y+z=1$
$2w+x+3y+z=7$
$2w+2x+y+2z=7$
consider $z$ fixed:
1)$w+x+y=1-z$
2)$2w+x+3y=7-z$
3)$2w+2x+y=7-2z$
with substract $2$ from $1$ we will have :$w+2y=6$
with substract $3$ from $-2(2)$ we will have : $-2w-5y=-7$
and so $w=16$ and $y=-5$
and with attention to (1) we will have
$x=1-z-16+5$
that $z$ is free variable (can take any value)
Suppose we have two linear equations,
$$y_1=m_1x+b_1$$
and
$$y_2=m_2x+b_2.$$
If we set $y_1=y_2$,then $m_1x+b_1=m_2x+b_2$. We then solve for $x$ and substitute that value into either equation to get the corresponding $y$ value. The result is the single intersection point of these two straight lines. It will exist provided $m_1 \neq m_2$. I have just made an online interactive plot for you to play with. Sometimes this visual representation is what helps a concept like this to be truly understood.
Try simultaneously solving various equations like these while at the same time dragging the sliders to the appropriate values in this demonstration I have constructed. I think with this, clarity will eventually come to you.
An answer to the actual question:
If we simply add the equations side by side, we get a third equation that has slope equal to the sum of slopes, and $y$ intercept equal to the sum of $y$ intercepts. You can see this by plotting $y_1+y_2$ in the demonstration. This would be a third line that does not necessarily pass through the intersect of the two equations. In order to find the solution to a system of equations, that being the point at which both equations are simultaneously true, we need to set equals to equals, ie. $x$ equal to $x$ or $y$ equal to $y$. In this manner we find the solution to the system.
I have two more plots to show you plot 1 is the first, and plot 2 is the other. I think I understand precisely why you are asking this question now after reflecting on what you must be studying. It has to do with operations on systems of equalities, and how after multiplying one equality by a constant, then adding it to the other, you arrive at an equation in one variable, and thus arrive at a solution. You see that the sum intersects when a scalar multiple of one equality is added to the other. If you can manage to add a scalar multiple of one equation to the other in such a way that one of the variables vanishes, then you will have that third equation, it being the sum of two equations (not the originals), and that third equation will indeed intersect at the intersection of the first two equations. Best of clarity and fortune in your studies. :)
Best Answer
Keep in mind that both sides of the equations are the same (that's the point of an equation). From basic algebra we know that you can manipulate an equation by adding or subtracting the same value on both sides, so if you have \begin{align} x+y &= 5, \tag{1}\\ 2x+y &= 8, \tag{2} \end{align} You can subtract the left hand side (LHS) of $(1)$ from the LHS of $(2)$, while subtracting the right hand side (RHS) of $(1)$ from the RHS of $(2)$ since the LHS and the RHS of each equation are the exact same thing. It's like you're subtracting $5$ from both sides, since $x+y$ really is equal to $5$. Hence we write $(2)$ as \begin{align} 2x+y &= 8\\ 2x + y - x - y &= 8 - 5\\ x &= 3, \end{align} which (since $x+y = 5$) implies $y$ must equal $2$.