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Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks.\newlineMcCoy's Bakery sold one customer 11 dozen chocolate cookies for $7\$7. The bakery also sold another customer 44 dozen chocolate cookies and 88 dozen oatmeal cookies for $76\$76. How much do the cookies cost?\newlineA dozen chocolate cookies cost $\$_____, and a dozen oatmeal cookies cost $\$_____.

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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.\newlineMcCoy's Bakery sold one customer 11 dozen chocolate cookies for $7\$7. The bakery also sold another customer 44 dozen chocolate cookies and 88 dozen oatmeal cookies for $76\$76. How much do the cookies cost?\newlineA dozen chocolate cookies cost $\$_____, and a dozen oatmeal cookies cost $\$_____.
  1. Define Variables: Let xx be the cost per dozen of chocolate cookies and yy be the cost per dozen of oatmeal cookies.\newlineFor the first customer: 11 dozen chocolate cookies = $7\$7.\newlineEquation: 1x+0y=71x + 0y = 7
  2. First Customer Equation: For the second customer: 44 dozen chocolate cookies and 88 dozen oatmeal cookies = $76\$76. Equation: 4x+8y=764x + 8y = 76
  3. 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}
  4. Create Augmented Matrix: Use row operations to find the reduced row echelon form.\newlineFirst, multiply the first row by −4-4 and add it to the second row to eliminate the xx-term in the second equation.\newline−4×∣1amp;0 7amp;∣+∣4amp;8 76amp;∣=∣4amp;8 76amp;∣+∣−4amp;0 −28amp;∣=∣0amp;8 48amp;∣-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}
  5. Reduce to Echelon Form: Now we have the new system of equations represented by the matrix:\newline(1amp;0amp;∣amp;7 0amp;8amp;∣amp;48)\begin{pmatrix} 1 & 0 & \vert & 7 \ 0 & 8 & \vert & 48 \end{pmatrix}
  6. Solve for yy: Divide the second row by 88 to solve for yy.\newline0amp;8amp;∣amp;48\begin{matrix} 0 & 8 & | & 48 \end{matrix} / 88 = 0amp;1amp;∣amp;6\begin{matrix} 0 & 1 & | & 6 \end{matrix}\newlineNow we have y=6y = 6.
  7. Solve for x: Substitute y=6y = 6 into the first equation to solve for xx.\newline1x+0(6)=71x + 0(6) = 7\newlinex=7x = 7
  8. Final Cost Solution: We have found that x=7x = 7 and y=6y = 6. A dozen chocolate cookies cost $7\$7, and a dozen oatmeal cookies cost $6\$6.

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