ALGEBRA QUESTION! need help ASAP!

Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $15 and same-day tickets cost $35. For one performance, there were 45 tickets sold in all, and the total amount paid for them was $1275. How many tickets of each type were sold?

Answer 1

a + s = 45

15a + 35s = 1275

a = 45 - s

15(45 - s) + 35s = 1275

The rest is on you. Go!

Answer 2

So this is a systems of equations problem, therefore we shall set it up like that.

Assign the variables:

1. a = advance tickets
2. s = same-day tickets

Now let's build the system of equations

Equation 1: Total number of tickets sold is 45 tickets

a + s = 45

Equation 2: Total value of the tickets is $1275 where advance tickets costs $15 per ticket and same-day is $35 per ticket

15a + 35s = 1275

Now we have our equations:

15a + 35s = 1275

a + s = 45

Let's use substitution method

a = 45 - s

Therefore (scroll down for step by step explanation):

1. Step 1: 15 (45 - s) + 35s = 1275
2. Step 2 : 675 - 15s + 35s = 1275
3. Step 3 : 20s = 600
4. Step 4 : s = 30
5. Step 5 : a = 45 - 30
6. Step 6 : a = 15

Therefore 15 advance tickets were sold and 30 same day tickets were solved

Step-by-Step Explanation:

1. Step 1: substitute the 45 - s into a in the other equation
2. Step 2: Distribute the 15 into the 45 - s
3. Step 3: Combine like terms by 35s - 15s = 20s and 1275 - 675 = 600
4. Step 4: divide 20 from each side to isolate the variable to get same day ticket = 30 tickets sold
5. Step 5: Substitute our new value for s into a = 45 - s to get a = 45 - 30
6. Step 6: Solve for a to get a = 15 advance tickets.