## Precalculus (6th Edition) Blitzer

a. Step 1. Assume the amounts of Food-1, Food-2, and Food-3 are $x,y,z$ respectively. We can set up a system of equations using the desired quantities of each vitamins: $\begin{cases} 20x+30y+10z=220 \\ 20x+10y+10z=180 \\ 10x+10y+30z=340 \end{cases}$ Simplify and rearrange the equations: $\begin{cases} x+y+3z=34\\ 2x+3y+z=22 \\ 2x+y+z=18 \end{cases}$ Step 2. If $x=0$, we have $\begin{cases} y+3z=34\\ 3y+z=22 \\ y+z=18 \end{cases}$ Step 3. Write a matrix of the system and perform row operations: $\begin{bmatrix} 0 & 1 & 3 & | & 34 \\ 0 & 3 & 1 & | & 22 \\ 0 & 1 & 1 & | & 18 \end{bmatrix} \begin{array} ..\\3R1-R2\to R2\\R1-R3\to R3 \end{array}$ $\begin{bmatrix} 0 & 1 & 3 & | & 34 \\ 0 & 0 & 8 & | & 80 \\ 0 & 0 & 2 & | & 6 \end{bmatrix} \begin{array} ..\\..\\R2-4R3\to R3 \end{array}$ $\begin{bmatrix} 0 & 1 & 3 & | & 34 \\ 0 & 0 & 8 & | & 80 \\ 0 & 0 & 0 & | & 56 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ Step 4. The last row gives $0=56$, which is not possible. Thus we have an inconsistent system and there is no solution. In other words, the dietary requirements cannot be met without Food-1. b. Start from the system in Step-1 and remove the equation for Vita-A (220mg). Make a matrix and perform row operations: $\begin{bmatrix} 1 & 1 & 3 & | & 34 \\ 2 & 1 & 1 & | & 18 \end{bmatrix} \begin{array} .\\2R1-R2\to R2 \end{array}$ $\begin{bmatrix} 1 & 1 & 3 & | & 34 \\ 0 & 1 & 5 & | & 50 \end{bmatrix} \begin{array} .\\.. \end{array}$ We have a dependent system. Let $z=t\geq0$; then $y+5t=50, y=50-5t,t\leq10$ and $x+y+3t=34, x=2t-16,t\geq8$. Thus $8\leq t\leq 10$. Let $t=8$; we have solution as $(0,10,8)$. Let $t=10$; we have $(4,0,10)$ as example solutions.