Question

three families visited the local fair over a holiday weekend: - the rawlins family spent $51 on admission for 2 children and 3 adults, plus 50 carnival tickets. - the alvarez family spent $90 on admission for 4 children and 6 adults, plus 100 carnival tickets. - the talbot family spent $78 on admission for 2 children and 5 adults, plus 80 carnival tickets. which matrix equation can be solved to find the cost of admission for children, x, the cost of admission for adults, y, and the cost of each ticket, z? a. \\(\

$$\begin{bmatrix}2 & 3 & 50 \\\\ 4 & 6 & 100 \\\\ 2 & 5 & 80\\end{bmatrix}$$

\

$$\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}$$

= \

$$\begin{bmatrix}78 \\\\ 90 \\\\ 51\\end{bmatrix}$$

\\) b. \\(\

$$\begin{bmatrix}2 & 3 & 50 \\\\ 4 & 6 & 100 \\\\ 2 & 5 & 80\\end{bmatrix}$$

\

$$\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}$$

= \

$$\begin{bmatrix}51 \\\\ 90 \\\\ 78\\end{bmatrix}$$

\\) c. \\(\

$$\begin{bmatrix}2 & 4 & 2 \\\\ 3 & 6 & 5 \\\\ 50 & 100 & 80\\end{bmatrix}$$

\

$$\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}$$

= \

$$\begin{bmatrix}78 \\\\ 90 \\\\ 51\\end{bmatrix}$$

\\) d. \\(\

$$\begin{bmatrix}2 & 4 & 2 \\\\ 3 & 6 & 5 \\\\ 50 & 100 & 80\\end{bmatrix}$$

\

$$\begin{bmatrix}x \\\\ y \\\\ z\\end{bmatrix}$$

= \

$$\begin{bmatrix}51 \\\\ 90 \\\\ 78\\end{bmatrix}$$

\\)

Explanation:

Step1: Translate Rawlins' spending

$2x + 3y + 50z = 51$

Step2: Translate Alvarez' spending

$4x + 6y + 100z = 90$

Step3: Translate Talbot' spending

$2x + 5y + 80z = 78$

Step4: Match to matrix form

The coefficient matrix rows are $[2,3,50]$, $[4,6,100]$, $[2,5,80]$; the constant vector is

$$\begin{bmatrix}51\\90\\78\end{bmatrix}$$

.

Answer:

B.

$$\begin{bmatrix} 2 & 3 & 50 \\ 4 & 6 & 100 \\ 2 & 5 & 80 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 51 \\ 90 \\ 78 \end{bmatrix}$$