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29 July, 02:24

There are 300 people at a play. Some paid $5.00 for reserved seats; others paid $3.00 for general admission seats. Total receipts were $1080. How many tickets of each kind were sold?

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  1. 29 July, 05:24
    0
    R+g=300, r=300-g

    5r+3g=1080, using r found above in this you get:

    5 (300-g) + 3g=1080

    1500-5g+3g=1080

    1500-2g=1080

    -2g=-420

    g=210, since r=300-g

    r=90

    So 90 reserved and 310 general tickets were sold.
  2. 29 July, 06:16
    0
    What we will have to do in this case is probably create a system of equations.

    First, assign the variables to what we're trying to find.

    Let x = the number of general admission tickets sold

    Let y = the number of reserved tickets sold

    Then x + y = the total number of tickets sold

    Then 3x + 5y = how much money was made

    Here is our system of equations:

    x + y = 300

    3x + 5y = 1080

    Let's first solve for x. That means will have to cancel out the y variables in both of the equations when we add them up together.

    Multiply the top equation by - 5.

    -5 (x + y) = 300 * - 3 = - 5x - 5y = - 1500

    Now we have the system of equations.

    -5x - 5y = - 1500

    3x + 5y = 1080

    Add them up together to get one equation.

    When you get the sum of the left-hand side, the y's cancel out.

    -5x - 5y + 3x + 5y = - 2x + 0y or just - 2x

    Now add up the right-hand side.

    -1500 + 1080 = - 420

    Now we have the equation - 2x = - 420. Solve for x.

    Divide both sides by - 2.

    -2x / - 2 = - 420 / - 2

    x = 210

    Now we know how many general admission tickets were sold.

    But what about the reserved tickets? Simply replace the x variable with the value of x in one of the equations.

    I'll pick x + y = 300 because it's simpler.

    Replace x with the value of x.

    210 + y = 300

    Solve for y.

    Subtract both sides by 210.

    210 + y - 210 = 300 - 210

    y = 90

    So, 210 general admission tickets were sold and 90 reserved tickets were sold.
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