Simultaneously solving of equations I am trying to refresh some math skills and I am struggling over the following problem. I tried to solve it with the help of a number of sources (i.e. http://www.idomaths.com/simeq.php), but I haven't succeeded yet. 
I would like to solve the equations below simultaneously by either elimination or substitution but I cannot get my head around it. Is there anybody who is able to help me?
$w = 0.080w + 0.632x + 0.264y + 0.080z$
$x = 0.184w + 0.368x + 0.368y + 0.184z$
$y = 0.368w + 0.368y + 0.368z$
$z = 0.368w + 0.368z$
$1 = w + x + y + z$
The answers should be:
$w = 0.286$
$x = 0.285$
$y = 0.263$
$z = 0.166$
This is an example taken from the book "Introduction to Operations Research" by Hillier and Lieberman on page $739$ ($9^{th}$ edition). 
Appreciate your assistance. 
 A: Using, $1 = w+x+y+z$, and solving for each variable and then substituting back into the original system, we can rewrite the system as follows (there are other approaches too).
We have: $w = 1 - x -y - z$
Substitute this into the first equation, we have:
$1 - x - y - z = 0.080w + 0.632x + 0.264y + 0.080z \rightarrow 0.08 w + 1.632x + 1.264y +1.08 z = 1$
Repeating the process, but for the second equation, we have: $x = 1 -w -y - z$
Substituting this into the second equation, we have:
$1 - w - y - z = 0.184w + 0.368x + 0.368y + 0.184z \rightarrow 1.184 w + 0.368x + 1.368y +1.184 z = 1$
Repeat for the next two equations, write that in matrix form, and then do the row reduced reduced echelon form as shown below.
$$\begin{bmatrix}0.08 & 1.632 & 1.264 & 1.08 & 1\\1.184 & 0.368 & 1.368 & 1.184 & 1\\1.368 & 1 & 0.368 & 1.368 & 1\\1.368 & 1 & 1 & 0.368 & 1\end{bmatrix}$$
Using Row-Reduced-Echelon-Form yields:
$$\begin{bmatrix}1 & 0 & 0 & 0 & 0.285654\\
0 & 1 & 0 & 0 & 0.284835\\
0 & 0 & 1 & 0 & 0.263181\\
0 & 0 & 0 & 1 & 0.16633\end{bmatrix}$$
Of course, this yields the result you provided.
Is that clear?
A: The first step is to re-arrange the five equations into standard form:
$0 = -0.920w + 0.632x + 0.264y + 0.080z$
$0 = 0.184w  - 0.632x + 0.368y + 0.184z$
$0 = 0.368w  +  0.0 x  - 0.632y + 0.368z$
$0 = 0.368w  +  0.0 x +  0.0y   - 0.632z$
$1 = 1.0w + 1.0x + 1.0y +1.0 z $
To solve this set of equations, first recognize that system appears to be over-determined:  there are five linear equations in four unknowns.  However, the first four equations are not linearly independent.  The $4\times4$ determinant of the coefficients is zero.
Therefore, to solve these equations, simply ignore any one of the first four equations, include the fifth (as the new fourth) and solve the set of four linearly independent linear equations in four unknowns by any convenient method.
