Question

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is

$35
. For one performance,
40
advance tickets and
30
same-day tickets were sold. The total amount paid for the tickets was
$1250
. What was the price of each kind of ticket?

Answer 1

Let's set up variables and equations for things that we know from the question, with a standing for advance tickets, and s standing for same day tickets.


`40a + 30s = 1250`

`a + s = 35`

Let's solve it through substitution. Isolate either variable, a or s. I'll choose to isolate a. Subtract s from both sides.


`a = -s + 35`

Plug the new knowledge into the first equation, since we now know the value of a


`40(-s +35) + 30s = 1250`.

Distribute the 40 to the -s and 35


`-40s + 1400 + 30s = 1250`

Combine the negative and positive s


`-10s + 1400 = 1250`

Subtract 1400 from both sides.


`-10s = -150`

Divide both sides by -10


`s = 15`

The total price of an advanced ticket and same-day ticket is 35, and we know the value of a same-day ticket, so the price of an advance ticket must be 20.

The final answers are:
advanced ticket = 20
same day ticket = 15

Answer 2

`Advance=x`


`Same.day=y`


`x+y=35` so
`x=35-y`


`40x+30y=1250`


`40(35-y)+30y=1250`


`-10y=-150`


`y=15`


`x+15=35`


`x=20`

Cost for Advance ticket: $20
Cost for Same-day ticket: $15