Solving independent linear equations 
\begin{align}
&{-}2y+2z-1=0 \tag{4} \\[4px]
&{-}2x+4y-2z-2=0 \tag{5} \\[4px]
&\phantom{-2}x-y+3/2=0 \tag{6}
\end{align}
Equation (6) is the sum of (4) and (5). There are only two independent equations.
  Putting $z=0$ in (5) and (6) and solving for x and y, we have 
  \begin{align}
x&=-2 \\[4px]
y&=-1/2
\end{align}



*

*equation (6) is the sum of (4) and (5): OK, I see it

*There are only two independent equations: I didn't get; what does this sentence mean?

*Putting $z=0$ in (5) and (6): why putting z=0 in equation (5) and (6)?
Please help

You can also think of as follows.


*

*A single variable linear equation represents a single point.

*A two-variable linear equation represents a line. If there are two two-variable linear equations, their solution is the intersection point of them as long as both equations are independent.

*A three-variable linear equation represent a plane. If there are 3 three-variable equations, their solution is the intersection point of them as long as they are independent. 
So if you have two 3-variable linear equations (or three 3-variable linear equations in which two of them are dependent), you cannot get a single point of intersection but a collection of points which are on a line, for example, $x+y+5/2=0$ in your case.
