Meal tickets at a large school cafeteria cost $4 for children and $9 for adults. On a certain day, if 14 times as many children's tickets were bought as adult meal tickets, and the total revenue
from selling all tickets was $2,584, about how many children bought meal tickets?

Let's define:

C = number of children tickets bought

A = number of adult tickets bought.

We know that "On a certain day,  if 14 times as many children's tickets were bought as adult meal tickets"

Then we have:

C = 14*A

And we also know that:

"the total revenue  from selling all tickets was $2,584"

Then:

C*$4 + A*$9 = $2,584

Then we have a system of equations:

C = 14*A

C*$4 + A*$9 = $2,584

The first step to solve these is to isolate one of the variables in one of the equations, and then replacing it in the other equation. We can already see that C is already isolated in the first equation, then we can just replace it in the second one to get:

(14*A)*$4 + A*$9 = $2,584

A*$56 + A*$9 = $2,584

A*$65 = $2,584

A = $2,584/$65 = 39.75

Here we should have a whole number, (So one of the numbers given may be incorrect) This is because we can not have a 0.75 of a ticket sold.

In this case, we know that:

C = A*14

C = 39.75*14 = 546.5

We could round this up to get a whole number; 547

About 547 children tickets were sold.

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph

sasasadaassasasasa opk ok you know