math

drama club sold adult tickets for $8 and kid tickets for $5. for the final performance a total of 226 tickets were sold for a total of $1670. how many adult tickets were sold for the performance?

X = # of adult tickets
Y = # of kid tickets

X + Y = 226
8X + 5Y = 1670

Solve those two equations in two unknowns. One way would be to use substitution. (Insert 226 - X in place of Y in the second equation).

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  1. Let x = adult's tickets
    y = children's tickets
    Equation Insert x value in
    1. x + y = 226 first equation
    180 + y=226
    2. 8x + 5y = 1670 y=226-180
    By substitution, y=46
    x=180,y=46 Ans.
    y = 226-226-x
    2. 8x + 5(226-x)=1670 multiply
    8x + 1130 - 5x=1670 cancellation
    8x-5x=1670-1130
    3x=540
    x=180 continue above:

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    2. 👎
  2. Let x = adult's tickets
    y = children's tickets
    Equation Insert x value in
    1. x + y = 226 first equation
    180 + y=226
    2. 8x + 5y = 1670 y=226-180
    By substitution, y=46
    x=180,y=46 Ans.
    y = 226-226-x
    2. 8x + 5(226-x)=1670 multiply
    8x + 1130 - 5x=1670 cancellation
    8x-5x=1670-1130
    3x=540 divide
    x=180 continue above:

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    2. 👎
  3. oof

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