A customer bought 8 video cassettes and 6 chocolates at a total cost of \( \$ 86 \). Another customer bought 8 videc cassettes and 4 chocolates at a total cost of \( \$ 84 \). Find the cost of a video cassette and a chocolate.

Let's denote the cost of a video cassette as \( x \) dollars and the cost of a chocolate as \( y \) dollars. From the given information, we can set up the following system of equations: 1. For the first customer: \( 8x + 6y = 86 \) 2. For the second customer: \( 8x + 4y = 84 \) We can solve this system of equations to find the values of \( x \) and \( y \), which represent the cost of a video cassette and a chocolate, respectively. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}8x+6y=86\\8x+4y=84\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{43-3y}{4}\\8x+4y=84\end{array}\right.\) - step2: Substitute the value of \(x:\) \(8\times \frac{43-3y}{4}+4y=84\) - step3: Simplify: \(86-2y=84\) - step4: Move the constant to the right side: \(-2y=84-86\) - step5: Subtract the numbers: \(-2y=-2\) - step6: Change the signs: \(2y=2\) - step7: Divide both sides: \(\frac{2y}{2}=\frac{2}{2}\) - step8: Divide the numbers: \(y=1\) - step9: Substitute the value of \(y:\) \(x=\frac{43-3\times 1}{4}\) - step10: Simplify: \(x=10\) - step11: Calculate: \(\left\{ \begin{array}{l}x=10\\y=1\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=10\\y=1\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(10,1\right)\) The solution to the system of equations is \( x = 10 \) and \( y = 1 \). Therefore, the cost of a video cassette is $10 and the cost of a chocolate is $1.