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 $65 . For one performance 25, advance tickets and 35 same-day tickets were sold. The total amount paid for the tickets was $1975. What was the price of each kind of ticket?

Answer 1

Advance tickets are $30 and same-day tickets are 35.

Explanation:

We are going to set up a system of equations to solve for each type of ticket. First, let's set up some parameters. We are going to call advance tickets a, and same-day tickets b.

With the info given we can see that together a single ticket of each equates to 65, so:
a + b = 65, our first equation

We also see 25 advance and 35 same-day sold for a total of 1975, so:
25a + 35b = 1975, our second equation

We now put them one atop the other
a + b = 65
25a + 35b = 1975

Next, we find a least common factor so that we can cancel out one of our variables and solve for the other. I make habit of choosing the smallest number in cases like this. We are going to multiply our first equation by -25. So now we have:

-25a - 25b = -1625
25a + 35b = 1975

Combine like-terms so we can solve for b:
(25a - 25a) + (35b - 25b) = (1975 - 1625)

which becomes
0a + 10b = 350 or 10b = 350

divide both sides by 10 to isolate variable b:

(10b/10) = (350/10)

which becomes

b = 35

We can now plug this back into that first equation to solve for variable a:
a + b = 65
a + 35 = 65

Subtract 35 from both sides to isolate variable a:

a + 35 - 35 = 65 - 35
a + 0 = (65 - 35) or a = (65 - 35)
a = 30

There you have it. Advance tickets are 30 bucks and same-days are 35.