A church luncheon made $\backslash (842.$ Adult tickets cost $\backslash )10$ each and children’s tickets cost $\$6$ each. The number of children was $12$ more than twice the number of adults. How many of each ticket were sold?

The number of children was $12$more than twice the number of adults and adult ticket cost$\$10$ each and children's tickets cost$\$6$each and a church luncheon made$\$842.$We have to find the number of each ticket were sold.

Let xbe the number of adult tickets.

Now, the number of children is $12$more than the number of adults.

So, the number of children's tickets is $2x+12.$

Thus, the table is

The total value of adult tickets and children's tickets are$\$842.$

Write the equation by adding the total value of all the types of coins.

So,$10\left(x\right)+6\left(2x+12\right)=842$

By proceeding with the equation we get,

$$\begin{gathered} 10\left(x\right)+6\left(2x+12\right)=842 \\ 10x+12x+72=842 \\ 22x=842-72 \\ 22x=770 \\ x=35 \end{gathered}$$

Thus, the number of adult tickets are$35$and children's tickets are$2\left(35\right)+12=82.$

To check the answer replace xby $35$into the original equation.

$$\begin{gathered} 10\left(35\right)+6\left(2\left(35\right)+12\right)=842 \\ 350++6\left(82\right)=842 \\ 350+492=842 \\ 842=842 \end{gathered}$$

It is true.

Thus, the number of adult tickets sold is$35$and children's ticket is$82.$