Answer
a. $\begin{cases} x+y=500 \\ y-z =-250 \\ x+z=750 \end{cases}$
b. $(750-t, t-250, t)$
c. $x=350, y=150$
Work Step by Step
a. Using the figure given in the exercise, we can set up a system of equations as:
$\begin{cases} 350+400=x+z \\ x+y=300+200 \\ z+450=y+700 \end{cases}$ or $\begin{cases} x+y=500 \\ y-z =-250 \\ x+z=750 \end{cases}$
b. Write the matrix and perform row operations:
$\begin{bmatrix} 1 & 1 & 0 & | & 500 \\ 0 & 1 & -1 & | &-250 \\ 1 & 0 & 1 & | & 750 \end{bmatrix}\begin{array} .. \\..\\ R1-R3\to R3 \end{array}$
$\begin{bmatrix} 1 & 1 & 0 & | & 500 \\ 0 & 1 & -1 & | &-250 \\ 0 & 1 & -1 & | & -250 \end{bmatrix}\begin{array} .. \\..\\ R3-R2 \to R3 \end{array}$
$\begin{bmatrix} 1 & 1 & 0 & | & 500 \\ 0 & 1 & -1 & | &-250 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}\begin{array} .. \\..\\ .. \end{array}$
We have a dependent system. Let $z=t\geq0$; then
$y=t-250, x=750-t, 250\leq t \leq 750$
Thus the solution set is
$(750-t, t-250, t)$ and $ 250\leq t \leq 750$
c. Let $z=400$. We have $x=350, y=150$ (cars per hour).
