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:

1. 👍
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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2. 👎

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