Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. $\newline$ McCoy's Bakery sold one customer $1$ dozen chocolate cookies for $\$7$ . The bakery also sold another customer $4$ dozen chocolate cookies and $8$ dozen oatmeal cookies for $\$76$ . How much do the cookies cost? $\newline$ A dozen chocolate cookies cost $\$$ _, and a dozen oatmeal cookies cost $\$$ _.

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Solve a system of equations using substitution: 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

McCoy's Bakery sold one customer

1

dozen chocolate cookies for

\$7

. The bakery also sold another customer

4

dozen chocolate cookies and

8

dozen oatmeal cookies for

\$76

. How much do the cookies cost?

\newline

A dozen chocolate cookies cost

\$

_____, and a dozen oatmeal cookies cost

\$

_____.

Define Variables: Let $x$ be the cost per dozen of chocolate cookies and $y$ be the cost per dozen of oatmeal cookies. $\newline$ For the first customer: $1$ dozen chocolate cookies = $\$7$ . $\newline$ Equation: $1x + 0y = 7$
First Customer Equation: For the second customer: $4$ dozen chocolate cookies and $8$ dozen oatmeal cookies = $\$76$ . Equation: $4x + 8y = 76$
Second Customer Equation: Create an augmented matrix to represent the system of equations: $\newline$ \begin{array}{cc|c} 1 & 0 & 7 \ 4 & 8 & 76 \end{array}
Create Augmented Matrix: Use row operations to find the reduced row echelon form. $\newline$ First, multiply the first row by $-4$ and add it to the second row to eliminate the $x$ -term in the second equation. $\newline$ $-4 \times \begin{vmatrix} 1 & 0 \ 7 & \end{vmatrix} + \begin{vmatrix} 4 & 8 \ 76 & \end{vmatrix} = \begin{vmatrix} 4 & 8 \ 76 & \end{vmatrix} + \begin{vmatrix} -4 & 0 \ -28 & \end{vmatrix} = \begin{vmatrix} 0 & 8 \ 48 & \end{vmatrix}$
Reduce to Echelon Form: Now we have the new system of equations represented by the matrix: $\newline$ $\begin{pmatrix} 1 & 0 & \vert & 7 \ 0 & 8 & \vert & 48 \end{pmatrix}$
Solve for $y$ : Divide the second row by $8$ to solve for $y$ . $\newline$ $\begin{matrix} 0 & 8 & | & 48 \end{matrix}$ / $8$ = $\begin{matrix} 0 & 1 & | & 6 \end{matrix}$ $\newline$ Now we have $y = 6$ .
Solve for x: Substitute $y = 6$ into the first equation to solve for $x$ . $\newline$ $1x + 0(6) = 7$ $\newline$ $x = 7$
Final Cost Solution: We have found that $x = 7$ and $y = 6$ . A dozen chocolate cookies cost $\$7$ , and a dozen oatmeal cookies cost $\$6$ .