Respuesta :
Using a system of equations, it is found that:
- For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
- For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
- For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.
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The variables of the system are:
- x, which is the number of $29 tickets sold.
- y, which is the number of $34 tickets sold.
Total of 8,000 seats, all can be sold, thus:
[tex]x + y = 8000 \rightarrow x = 8000 - y[/tex]
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For a profit of $255,000, we have that:
[tex]29x + 34y = 255000[/tex]
[tex]29(8000 - y) + 34y = 255000[/tex]
[tex]5y = 23000[/tex]
[tex]y = \frac{23000}{5}[/tex]
[tex]y = 4600[/tex]
[tex]x = 8000 - 4600 = 3400[/tex]
For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
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For a profit of $271,000, we have that:
[tex]29x + 34y = 271000[/tex]
[tex]29(8000 - y) + 34y = 271000[/tex]
[tex]5y = 39000[/tex]
[tex]y = \frac{39000}{5}[/tex]
[tex]y = 7800[/tex]
[tex]x = 8000 - 7800= 200[/tex]
For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
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For a profit of $235,000, we have that:
[tex]29x + 34y = 235000[/tex]
[tex]29(8000 - y) + 34y = 235000[/tex]
[tex]5y = 3000[/tex]
[tex]y = \frac{3000}{5}[/tex]
[tex]y = 600[/tex]
[tex]x = 8000 - 600 = 7400[/tex]
For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.
A similar problem is given at https://brainly.com/question/22826010
