Substitution word problem

Question

This question is very confusing to me. I understand can do normal substitution but word problems are more difficult.

The members of the city cultural center have decided to put on a play once a night for a week. Their auditorium holds 600 people. By selling​ tickets, the members would like to raise $3,900 every night to cover all expenses. Let d represent the number of adult tickets sold at ​$8.50. Let s represent the number of student tickets sold at

​$5.50 each. If all 600 seats are filled for a​ performance, how many of each type of ticket must have been sold for the members to raise exactly $3,900​?

At one performance there were two times as many student tickets sold as adult tickets. If there were 300 tickets sold at that​ performance, how much below the goal of ​$3,900 did ticket sales​ fall?

Answer 1

Set up your system of equations like this:

d + s = 600

8.5d + 5.5s = 3,900

d = 600 - s (solving for d to substitute)

8.5(600 - s) + 5.5s = 3,900 (plug in known values)

5,100 - 8.5s + 5.5s = 3,900 (multiply)

-3s = -1,200 (isolate variable)

s = 400 (solve for s)

So there were 400 students and 200 adults every night.

For the separate performance with 300 tickets, divide both original sales in half.

200s x 5.5 = 1,100 in student tickets

100d x 8.5 = 850 in adult tickets

1,950 total sales

3,900 - 1,950 = $1,950 shortage