250 adult tickets sold
96 child tickets sold
1) We can solve this problem by resorting to a Linear System of Equations.
2) To do that, let's call a adult and c for the children. The first equation relates the costs and the revenue:
[tex]10a+5c=2980[/tex]And for the second equation, the total number of tickets.
[tex]a+c=346[/tex]3) So now, let's set this up and solve it by using the Elimination Method
[tex]\begin{gathered} 10a+5c=2980 \\ a+c=346\:\:\times(-5) \\ \\ 10a+5c=2980 \\ -5a-5c=-1730 \\ ----------- \\ 5a=1250 \\ \\ \frac{5a}{5}=\frac{1250}{5} \\ \\ a=250 \\ \\ \end{gathered}[/tex]To find c we can plug a=250 into the second equation
[tex]\begin{gathered} a+c=346 \\ \\ 250+c=346 \\ \\ -250+250+c=346-250 \\ \\ c=96 \end{gathered}[/tex]4) Thus, the answer is:
250 adult tickets sold
96 child ticket sold
