2) The school that Gabriella goes to is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 4 adult tickets and 4 child tickets for a total of \( \$ 76 \). The school took in \( \$ 94 \) on the second day by selling 6 adult tickets and 1 child tickets. Write a system of equations for this situation. Solve the system for extra credit.

To solve the system of equations for the situation described, we need to set up two equations based on the information provided. Let's denote: - \( A \) as the price of an adult ticket - \( C \) as the price of a child ticket From the first day of ticket sales, we know that the school sold 4 adult tickets and 4 child tickets for a total of $76. This can be represented by the equation: \[ 4A + 4C = 76 \] From the second day of ticket sales, we know that the school took in $94 by selling 6 adult tickets and 1 child ticket. This can be represented by the equation: \[ 6A + C = 94 \] Now, we have a system of two equations: \[ 4A + 4C = 76 \] \[ 6A + C = 94 \] To solve this system of equations, we can use the method of substitution or elimination. Let's use the elimination method to find the values of \( A \) and \( C \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4A+4C=76\\6A+C=94\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}4A+4C=76\\C=94-6A\end{array}\right.\) - step2: Substitute the value of \(C:\) \(4A+4\left(94-6A\right)=76\) - step3: Simplify: \(-20A+376=76\) - step4: Move the constant to the right side: \(-20A=76-376\) - step5: Subtract the numbers: \(-20A=-300\) - step6: Change the signs: \(20A=300\) - step7: Divide both sides: \(\frac{20A}{20}=\frac{300}{20}\) - step8: Divide the numbers: \(A=15\) - step9: Substitute the value of \(A:\) \(C=94-6\times 15\) - step10: Calculate: \(C=4\) - step11: Calculate: \(\left\{ \begin{array}{l}A=15\\C=4\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}A=15\\C=4\end{array}\right.\) - step13: Rewrite: \(\left(A,C\right) = \left(15,4\right)\) The solution to the system of equations is \( A = 15 \) and \( C = 4 \). Therefore, the price of an adult ticket is $15 and the price of a child ticket is $4.