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

$Advance=x$

$Same.day=y$

$x+y=35$ 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

Answer 2

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