Intermediate algebra

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 $50. For one performance, $30 advance tickets and $35 same-day tickets were sold. The total amount paid for the tickets was $1575. What was the price of each kind of ticket?

Answer 1

Can also use substitution method since a = 50 - s

30a + 35s = 1575 becomes 30(50 - s) + 35s = 1575

1500s - 30s + 35s = 1575 using distribution rule

5s = 75

s = $15 and a = 50 - 15 = $35

Check: 30(35) + 35(15) = 1050 + 525 = $1575

Answer 2

Suppose the variable "a" represents advance tickets and "s" represents same-day tickets.

The problem states that one advance ticket and one same-day ticket sold for $50. We can write this statement as an equation: a+s=50.

It also states that one day 30 advance tickets and 35 same-day tickets were sold, and the total amount for both types of tickets was $1,575. We can write this as 30a+35s=1575.

Now, we need to solve the system of equations. I believe that elimination would be easiest for these two equations. We can multiply the first equation by 30 so that the "a" variable cancels.

The first equation now becomes 30a+30s=1500.

Now, let's eliminate by subtracting both equations.

This gives us -5s=-75, so s=15.

We can solve for "a" by plugging in "s" to the original equation and solving for "a".

Substituting "s" gives us a+15=50, so "a" must equal 35.

Therefore, advance tickets cost $35 and same-day tickets cost $15. Hope this helps!

Answer 3

a + s = 50

30a + 35s = 1576

Can you use substittution, solve, and answer?