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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 $65. For
one performance, 40 advance tickets and 20 same-day tickets were sold. The total amount paid for the tickets was $1900. What was the price of each kind of
ticket?

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Answer:

advance tickets $30.00

same day tickets $35.00

Step-by-step explanation:

1. In this equation, you are trying to find two values so you must use system of equations. Your two equations would be:

40x +20y = 1900

x + y = 65

x represents the price of advance tickets and y represents the price of same day tickets.

2. Solve the equation by performing elimination using multiplication. Multiple one equation by a constant to get two equations that contain opposite terms.

40x + 20y = 1900

-40(x + y = 65) → -40x + -40y = -2600

3. Add the equations, eliminating one variable. (add -40x + -40y = -2600 to 40x + 20y = 1900)

   40x + 20y = 1900

+ -40x + -40y = -2600

Results in: -20y = -700

Solve for y by dviding both sides by -20.

y = 35

4. Substitute the solved value into one of the equations and solve for the remaining variable.

x + 35 = 65

Subtract 35 from both sides to solve for x.

x = 30

The advance tickets were $30.00 and the same day tickets were $35.00.

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