Question

there were 300 tickets $1600 the cost of the tickets was $8 per adult and $4 per child how many of each?

Answer 1

let \(a\) be the number of adult tickets at $8 each
let \(c\) be the number of child tickets at $4 each

We know that there are 300 tickets, so \[a+c=300\]

We know that the revenue from adult tickets is \(8a\)
We know that the revenue from child tickets is \(4c\)

We know the total ticket revenue is $1600, so
\[8a+4c=1600\]

That gives us two equations in two unknowns, which can be solved easily by substitution or elimination.