12. Admission to the gymnastics meet cost \( \$ 2.40 \) for adults and \( \$ 1.25 \) for students. If 540 tickets were sold for a total of \( \$ 934.90 \), how many tickets of each kind were sold?

To solve this problem, we can set up a system of equations based on the given information. Let's denote the number of adult tickets sold as \( A \) and the number of student tickets sold as \( S \). Given: - The cost of an adult ticket is \( \$2.40 \). - The cost of a student ticket is \( \$1.25 \). - A total of 540 tickets were sold. - The total amount collected from ticket sales is \( \$934.90 \). We can set up the following system of equations: 1. \( A + S = 540 \) (Total number of tickets sold) 2. \( 2.40A + 1.25S = 934.90 \) (Total amount collected from ticket sales) Now, we can solve this system of equations to find the number of adult and student tickets sold. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}A+S=540\\2.4A+1.25S=934.9\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}A=540-S\\2.4A+1.25S=934.9\end{array}\right.\) - step2: Substitute the value of \(A:\) \(2.4\left(540-S\right)+1.25S=934.9\) - step3: Simplify: \(1296-1.15S=934.9\) - step4: Move the constant to the right side: \(-1.15S=934.9-1296\) - step5: Subtract the numbers: \(-1.15S=-361.1\) - step6: Change the signs: \(1.15S=361.1\) - step7: Divide both sides: \(\frac{1.15S}{1.15}=\frac{361.1}{1.15}\) - step8: Divide the numbers: \(S=314\) - step9: Substitute the value of \(S:\) \(A=540-314\) - step10: Calculate: \(A=226\) - step11: Calculate: \(\left\{ \begin{array}{l}A=226\\S=314\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}A=226\\S=314\end{array}\right.\) - step13: Rewrite: \(\left(A,S\right) = \left(226,314\right)\) The solution to the system of equations is \( A = 226 \) and \( S = 314 \). Therefore, 226 adult tickets and 314 student tickets were sold.