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12 June, 12:19

Takeo and Manuel drove to a baseball game together. the tickets cost $10.50 each and parking was $6. popcorn cost $2.25 a box and each of the boys bought a box of popcorn. takeo bought one drink and manuel bought 2 drinks. the baseball game and the snacks cost the boys a total of $39.75. how much did each of the drinks cost

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Answers (2)
  1. 12 June, 15:02
    0
    Tickets: $10.50 x 2 = $21.00

    Parking: $6.00 x 1 = $6.00 they drove together so they only parked 1 car

    Popcorn: $2.25 x 2 = $4.50

    Drinks: D x 3 = 3D

    Tickets + Parking + Popcorn + Drinks = Total

    $21.00 + $6.00 + $4.50 + 3D = $39.75

    $31.50 + 3D = $39.75

    3D = $ 8.25

    D = $2.75

    Answer: $2.75 each drink
  2. 12 June, 15:11
    0
    Make variables for each of the coefficients in this problem. Tickets will be 10.50t, parking is 6, popcorn is 2.25p, and drinks are d. Make all of these add up to 39.75.

    10.50t + 6 + 2.25p + d = 39.75. Substitute the amount they bought for the variables, so now your equation will look like this: 10.50 (2) + 6 + 2.25 (2) + 3d = 39.75.

    Multiply 10.50 times 2 and 2.25 times 2.

    21 + 6 + 4.5 + 3d = 39.75, then add 21 plus 6 plus 4.5.

    31.5 + 3d = 39.75, then subtract 31.5 from both sides.

    3d = 8.25, now divide both sides by 3.

    d = 2.75. Each of the drinks cost $2.75.
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