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.
${\begin{cases} x + y + z = 2680 \\ 8 x + 10 y + 12 z = 300 \\ x + y + 5 z = 0 \end{cases}$
${\begin{cases} x + y + z = 300 \\ 8 x + 10 y + 12 z = 2680 \\ x + y - 5 z = 0 \end{cases}$
${\begin{cases} x + y + z = 300 \\ 8 x + 10 y + 12 z = 2680 \\ x + y + 5 z = 0 \end{cases}$
${\begin{cases} x + y + z = 2680 \\ 8 x + 10 y + 12 z = 300 \\ x + y - 5 z = 0 \end{cases}$
There were $◻$ tickets of $$ 8$ , $◻$ tickets of $$ 10$ , and $◻$ tickets of $$ 12$ sold

Answer

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Answer

Final Answer: There were $210$ tickets of $8, $40 $ t i c k e t s o f$ 10, and $50$ tickets of $12 sold.

Steps

Step 1 ：Let's denote the number of $8 t i c k e t s a s x,$ 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 $8 x + 10 y + 12 z = 2680$ .

Step 4 ：The combined number of $8 a n d$ 10 tickets sold was 5 times the number of $12 tickets sold, so we have the equation $x + y = 5 z$ .

Step 5 ：So, the correct system of linear equations is: ${\begin{cases} x + y + z = 300 \\ 8 x + 10 y + 12 z = 2680 \\ x + y = 5 z \end{cases}$

Step 6 ：Solving this system of equations, we find that $x = 210$ , $y = 40$ , and $z = 50$ .

Step 7 ：Final Answer: There were $210$ tickets of $8, $40 $ t i c k e t s o f$ 10, and $50$ tickets of $12 sold.