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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 :

[tex]Advance=x[/tex]

[tex]Same.day=y[/tex]

[tex]x+y=35[/tex] so [tex]x=35-y[/tex]

[tex]40x+30y=1250[/tex]

[tex]40(35-y)+30y=1250[/tex]

[tex]-10y=-150[/tex]

[tex]y=15[/tex]

[tex]x+15=35[/tex]

[tex]x=20[/tex]

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

[tex]40a + 30s = 1250[/tex] 
[tex]a + s = 35[/tex]

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

[tex]a = -s + 35[/tex]

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

[tex]40(-s +35) + 30s = 1250[/tex].

Distribute the 40 to the -s and 35

[tex]-40s + 1400 + 30s = 1250[/tex]

Combine the negative and positive s

[tex]-10s + 1400 = 1250[/tex]

Subtract 1400 from both sides.

[tex]-10s = -150[/tex]

Divide both sides by -10

[tex]s = 15 [/tex]

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


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