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Answer :
Answer:
[tex]\left\{\begin{array}{l}x+y\ge 20\\ \\10x+20y\le 280\end{array} \right.[/tex]
Step-by-step explanation:
Let x be the number of $10 tickets the radio station gives away and y is the number of $20 tickets which the radio station gives away.
Number of tickets:
At least 20, so
[tex]x+y\ge 20[/tex]
Cost:
x tickets for $10 each cost $10x
y tickets for $20 each cost $20y
Total cost:
$(10x+20y)
The total cost of all the tickets they give away can be no more than $280, so
[tex]10x+20y\le 280[/tex]
We get the system of two inequalities:
[tex]\left\{\begin{array}{l}x+y\ge 20\\ \\10x+20y\le 280\end{array} \right.[/tex]
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Final answer:
To solve the problem, set up a system of inequalities using x and y to represent the number of $10 and $20 tickets given away. Solve the system to get the possible values of x and y.
Explanation:
To solve this problem, let x be the number of $10 tickets and y be the number of $20 tickets given away. We can set up a system of inequalities to represent the given conditions:
10x + 20y ≥ 280 (total cost should be no more than $280)
x + y ≥ 20 (at least 20 tickets should be given away)
Solving this system of inequalities will give us the possible values of x and y that satisfy the conditions.
Learn more about Solving a system of inequalities here:
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