Use a system of linear equations in three variables to solve the following - problem.
At a college production of a play, 300 tickets were sold. The ticket prices were $\$ 8, \$ 10$, and $\$ 12$, and the total income from ticket sales was $\$ 2680$. How many tickets of each type were sold if the combined number of $\$ 8$ and $\$ 10$ tickets sold was 5 times the number of $\$ 12$ tickets sold?
Write a system of linear equations using the given information. Choose correct answer below.
$\left\{\begin{array}{l}x+y+z=2680 \\ 8 x+10 y+12 z=300 \\ x+y+5 z=0\end{array}\right.$
$\left\{\begin{array}{l}x+y+z=300 \\ 8 x+10 y+12 z=2680 \\ x+y-5 z=0\end{array}\right.$
$\left\{\begin{array}{l}x+y+z=300 \\ 8 x+10 y+12 z=2680 \\ x+y+5 z=0\end{array}\right.$
$\left\{\begin{array}{l}x+y+z=2680 \\ 8 x+10 y+12 z=300 \\ x+y-5 z=0\end{array}\right.$
There were $\square$ tickets of $\$ 8$, $\square$ tickets of $\$ 10$, and $\square$ tickets of $\$ 12$ sold
Answer
Final Answer: There were \(\boxed{210}\) tickets of $8, \(\boxed{40}\) tickets of $10, and \(\boxed{50}\) tickets of $12 sold.
Step 1 ：Let's denote the number of $8 tickets as x, $10 tickets as y, and $12 tickets as z.
Step 2 ：From the problem, we know that the total number of tickets sold is 300, so we have the equation \(x + y + z = 300\).
Step 3 ：The total income from ticket sales was $2680, so we have the equation \(8x + 10y + 12z = 2680\).
Step 4 ：The combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold, so we have the equation \(x + y = 5z\).
Step 5 ：So, the correct system of linear equations is: \[\left\{\begin{array}{l}x+y+z=300 \\ 8 x+10 y+12 z=2680 \\ x+y=5z\end{array}\right.\]
Step 6 ：Solving this system of equations, we find that \(x = 210\), \(y = 40\), and \(z = 50\).
Step 7 ：Final Answer: There were \(\boxed{210}\) tickets of $8, \(\boxed{40}\) tickets of $10, and \(\boxed{50}\) tickets of $12 sold.
