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Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks.\newlineTwo fifth grade classes are attending an amusement park as a field trip. Their teachers purchased the tickets for the class and the parent chaperones. Mrs. Randolph purchased 3030 child tickets and 2323 adult tickets, which cost a total of $1,359\$1,359. Mr. Kent purchased 11 adult ticket, paying a total of $33\$33. What are the ticket prices?\newlineThe price of a child ticket is $____\$\_\_\_\_ and the price of an adult ticket is $____\$\_\_\_\_.

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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.\newlineTwo fifth grade classes are attending an amusement park as a field trip. Their teachers purchased the tickets for the class and the parent chaperones. Mrs. Randolph purchased 3030 child tickets and 2323 adult tickets, which cost a total of $1,359\$1,359. Mr. Kent purchased 11 adult ticket, paying a total of $33\$33. What are the ticket prices?\newlineThe price of a child ticket is $____\$\_\_\_\_ and the price of an adult ticket is $____\$\_\_\_\_.
  1. Define Variables: Let xx be the price of a child ticket and yy be the price of an adult ticket.\newlineMrs. Randolph's purchase: 30x+23y=135930x + 23y = 1359\newlineMr. Kent's purchase: x+y=33x + y = 33
  2. Create Augmented Matrix: Write the system of equations as an augmented matrix:\newline\begin{array}{cc|c} 30 & 23 & 1359 \ 1 & 1 & 33 \end{array}
  3. Transform to Row-Echelon Form: Use row operations to transform the matrix into row-echelon form.\newlineFirst, swap the rows so the smaller coefficient of xx is in the top row.\newline\begin{align*}\(\newline&\begin{array}{cc|c}\newline1 & 1 & 33 (\newline\)30 & 23 & 1359\newline\end{array}\newline\end{align*}\)
  4. Eliminate Variable xx: Multiply the first row by 30-30 and add it to the second row to eliminate xx from the second row.\newline30×(1133)+(30231359)=(0733+1359)-30 \times (1 1 | 33) + (30 23 | 1359) = (0 -7 | -33 + 1359)\newlineThis results in the new second row: (071326)(0 -7 | 1326)
  5. Solve for Variable yy: Divide the second row by 7-7 to solve for yy.\newline(071326)/7=(01189)(0 -7 | 1326) / -7 = (0 1 | -189)\newlineThis gives us y=189y = -189, which is incorrect because ticket prices can't be negative.

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