Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ The administrative assistant at a software company often provides breakfast when there is a morning meeting. For last week's sales meeting, she purchased $8$ dozen doughnuts and $2$ dozen croissants, spending a total of $\$68$ . In preparation for yesterday's safety meeting, she spent $\$10$ on $1$ dozen croissants. Assuming she purchased the items at the same bakery both times, how much does a dozen of each cost? $\newline$ A dozen doughnuts costs $\$$ $\_\_\_\_$ , and a dozen croissants costs $\$$ $\_\_\_\_$ .

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Solve a system of equations using any method: word problems

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Q. Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks.

\newline

The administrative assistant at a software company often provides breakfast when there is a morning meeting. For last week's sales meeting, she purchased

8

dozen doughnuts and

2

dozen croissants, spending a total of

\$68

. In preparation for yesterday's safety meeting, she spent

\$10

1

dozen croissants. Assuming she purchased the items at the same bakery both times, how much does a dozen of each cost?

\newline

A dozen doughnuts costs

\$

\_\_\_\_

, and a dozen croissants costs

\$

\_\_\_\_

Formulate Equations: First equation based on $8$ dozen doughnuts and $2$ dozen croissants for $\$68$ . $8d + 2c = 68$
Create Augmented Matrix: Second equation based on $1$ dozen croissants for $\$10$ . $0d + 1c = 10$
Find Croissant Cost: Write the system of equations as an augmented matrix. $\newline$ \left[\begin{array}{cc|c}$\newline8 & 2 & 68 (\newline$0 & 1 & 10 $\newline$ \end{array}\right]\)
Calculate Doughnut Cost: Use row operations to find the cost of a dozen croissants. $\newline$ Since the second row is already solved for $c$ , we know that $c = 10$ .
Verify Solution: Substitute $c = 10$ into the first equation to find $d$ . $8d + 2(10) = 68$ $8d + 20 = 68$ $8d = 68 - 20$ $8d = 48$ $d = \frac{48}{8}$ $d = 6$
Verify Solution: Substitute $c = 10$ into the first equation to find $d$ .
$8d + 2(10) = 68$
$8d + 20 = 68$
$8d = 68 - 20$
$8d = 48$
$d = \frac{48}{8}$
$d = 6$ Check the solution by substituting $d$ and $c$ back into the original equations.
$d$ $0$
$d$ $1$
$d$ $2$
$d$ $3$
$d$ $4$
$d$ $5$