## Precalculus (6th Edition) Blitzer

dependent system; two example solutions: $(7, 3, 0)$ and $(7,0,6)$.
Step 1. Assume the hours for Product-A, Product-B, and Product-C, are $x,y,z$ respectively. We can set up a system of equations using the given conditions: $\begin{cases} 7x+6y+3z=67 \\ 2x+2y+z = 20 \end{cases}$ Step 2. Write a matrix of the system and perform the row operations: $\begin{bmatrix} 7 & 6 & 3 & | & 67 \\ 2 & 2 & 1 & | & 20 \end{bmatrix} \begin{array} ..\\7R2-2R1\to R2 \end{array}$ $\begin{bmatrix} 7 & 6 & 3 & | & 67 \\ 0 & 2 & 1 & | & 6 \end{bmatrix} \begin{array} ..\\.. \end{array}$ Step 3. As we have a dependent system, let $z=t\geq0$. We have $2y+t=6, y=\frac{6-t}{2}, t\leq6$, and $2x+2y+t=20, x=7$. Thus we have the solution set as $(7, \frac{6-t}{2}, t)$, with $0\leq t\leq 6$ Step 4. To get two examples, let $t=0$; we have the solution $(7, 3, 0)$. Let $t=6$; we have $(7,0,6)$.