Answer
a. $\begin{cases} w+z=380 \\ w+x=600 \\ x-y=170 \\ y-z=50 \end{cases}$
b. $(380-t,t+220,t+50,t)$
c. $w=330, x=270, y=100$
Work Step by Step
a. At each intersection, the input equals the output; we have:
$\begin{cases} w+z=200+180 \\ w+x=400+200 \\ y+200=x+30 \\ z+70=y+20 \end{cases}$ or $\begin{cases} w+z=380 \\ w+x=600 \\ x-y=170 \\ y-z=50 \end{cases}$
b. Set up the matrix based on the above equations:
$\begin{bmatrix} 1 & 0 & 0 & 1 & | & 380 \\ 1 & 1 & 0 & 0 & | & 600 \\ 0 & 1 & -1 & 0 & | & 170 \\ 0 & 0 & 1 & -1 & | & 50 \end{bmatrix} \begin{array} ..\\R2-R1\to R2\\..\\.. \end{array}$
$\begin{bmatrix} 1 & 0 & 0 & 1 & | & 380 \\ 0 & 1 & 0 & -1 & | & 220 \\ 0 & 1 & -1 & 0 & | & 170 \\ 0 & 0 & 1 & -1 & | & 50 \end{bmatrix} \begin{array} ..\\..\\R2-R3\to R3\\.. \end{array}$
$\begin{bmatrix} 1 & 0 & 0 & 1 & | & 380 \\ 0 & 1 & 0 & -1 & | & 220 \\ 0 & 0 & 1 & -1 & | & 50 \\ 0 & 0 & 1 & -1 & | & 50 \end{bmatrix} \begin{array} ..\\..\\..\\.. \end{array}$
As the law two rows are identical, the system is dependent. Letting $z=t$, we have $y=t+50, x=170+y=t+220, w=380-t$ and the solution set can be written as $(380-t,t+220,t+50,t)$
c. Let $z=t=50$; we have $w=330, x=270, y=100$
