Students are selling brownies and chocolate bars for a fundraiser. They sold 22 items and made \( \$ 76 \). brownie: \( \$ 3 \) chocolate bar: \( \$ 4 \)

To solve this problem, we can set up a system of equations based on the information provided. Let's denote the number of brownies sold as \( b \) and the number of chocolate bars sold as \( c \). Given: - Total number of items sold: \( b + c = 22 \) - Total amount of money made: \( 3b + 4c = 76 \) We can solve this system of equations to find the number of brownies and chocolate bars sold. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}b+c=22\\3b+4c=76\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}b=22-c\\3b+4c=76\end{array}\right.\) - step2: Substitute the value of \(b:\) \(3\left(22-c\right)+4c=76\) - step3: Simplify: \(66+c=76\) - step4: Move the constant to the right side: \(c=76-66\) - step5: Subtract the numbers: \(c=10\) - step6: Substitute the value of \(c:\) \(b=22-10\) - step7: Calculate: \(b=12\) - step8: Calculate: \(\left\{ \begin{array}{l}b=12\\c=10\end{array}\right.\) - step9: Check the solution: \(\left\{ \begin{array}{l}b=12\\c=10\end{array}\right.\) - step10: Rewrite: \(\left(b,c\right) = \left(12,10\right)\) The solution to the system of equations is \( b = 12 \) and \( c = 10 \). Therefore, the students sold 12 brownies and 10 chocolate bars, making a total of \( \$ 76 \).