Tickets to a show cost$10 in advance and $15 at the door. If 120 Tickets were sold for a total of $1390, how many of the tickets were bought in advance?

Let x = advance and y = at door.

x = 120 - y

10x + 15y = 1390

Substitute 120-y for x in second equation and solve for y. Insert that value into the first equation and solve for x. Check by inserting both values into the second equation.

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To solve this problem, we can use a system of equations.

Let's assume the number of tickets bought in advance is represented by 'x', and the number of tickets bought at the door is represented by 'y'.

Given that the total number of tickets sold is 120, we can write our first equation as:

x + y = 120 ... Equation 1

It's also given that the total revenue from ticket sales is $1390. Since each ticket bought in advance costs $10 and each ticket bought at the door costs $15, we can write our second equation as:

10x + 15y = 1390 ... Equation 2

Now we have a system of equations:

x + y = 120 ... Equation 1
10x + 15y = 1390 ... Equation 2

We can solve this system of equations using substitution or elimination to find the values of 'x' and 'y'. Let's use elimination:

Multiply Equation 1 by 10 to make the coefficients of 'x' in both equations the same:

10x + 10y = 1200 ... Equation 3

Now subtract Equation 3 from Equation 2:

(10x + 15y) - (10x + 10y) = 1390 - 1200

5y = 190

Divide both sides of the equation by 5:

y = 190/5 = 38

Now substitute the value of 'y' into Equation 1 to solve for 'x':

x + 38 = 120

x = 120 - 38 = 82

Therefore, 82 tickets were bought in advance.