A food truck sells sandwiches and drinks. One customer bought 4 sandwiches and 6 drinks for \( \$ 30 \). Another customer bought 3 sandwiches and 3 drinks for \( \$ 19.50 \). How much will a customer pay for 5 sandwiches and 4 drinks? A. \( \$ 28 \) B. \( \$ 30.50 \) C. \( \$ 36 \) D. \( \$ 49.50 \)

Let's denote the cost of one sandwich as \( s \) and the cost of one drink as \( d \). From the first customer's purchase: 4 sandwiches and 6 drinks cost $30, so we have the equation: \[ 4s + 6d = 30 \] From the second customer's purchase: 3 sandwiches and 3 drinks cost $19.50, so we have the equation: \[ 3s + 3d = 19.50 \] We can solve these two equations to find the values of \( s \) and \( d \), and then calculate the cost of 5 sandwiches and 4 drinks. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4s+6d=30\\3s+3d=19.5\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}d=\frac{15-2s}{3}\\3s+3d=19.5\end{array}\right.\) - step2: Substitute the value of \(d:\) \(3s+3\times \frac{15-2s}{3}=19.5\) - step3: Simplify: \(s+15=19.5\) - step4: Move the constant to the right side: \(s=19.5-15\) - step5: Subtract the numbers: \(s=4.5\) - step6: Substitute the value of \(s:\) \(d=\frac{15-2\times 4.5}{3}\) - step7: Simplify: \(d=2\) - step8: Calculate: \(\left\{ \begin{array}{l}d=2\\s=4.5\end{array}\right.\) - step9: Check the solution: \(\left\{ \begin{array}{l}d=2\\s=4.5\end{array}\right.\) - step10: Rewrite: \(\left(d,s\right) = \left(2,4.5\right)\) The solution to the system of equations is \( d = 2 \) and \( s = 4.5 \). Now, we can calculate the cost of 5 sandwiches and 4 drinks: \[ 5s + 4d = 5(4.5) + 4(2) = 22.5 + 8 = 30.5 \] Therefore, a customer will pay $30.50 for 5 sandwiches and 4 drinks. So, the answer is B. $30.50.