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Answer :
We can form two equations, let the price of a senior ticket be s and the price of a child ticket be c.
We have from day 1:
A: 3s + 9c = 75
And from day 2:
B: 8s + 5c = 67
Now we can rewrite A as:
A: 24s + 72c = 600
And can rewrite B as:
B: 24s + 15c = 201
Now A-B can be written as:
A-B: 57c = 399
So c = 7
Now substituting this back into A we get:
A: 3s + 63 = 75
A: 3s = 12
So s = 4
We have the price of a senior ticket is $4 and the price of a child ticket is $7
We have from day 1:
A: 3s + 9c = 75
And from day 2:
B: 8s + 5c = 67
Now we can rewrite A as:
A: 24s + 72c = 600
And can rewrite B as:
B: 24s + 15c = 201
Now A-B can be written as:
A-B: 57c = 399
So c = 7
Now substituting this back into A we get:
A: 3s + 63 = 75
A: 3s = 12
So s = 4
We have the price of a senior ticket is $4 and the price of a child ticket is $7
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Rewritten by : Barada
The price of one senior citizen ticket is $9 and the price of one child ticket is $4.
We have,
Let's assume the price of a senior citizen ticket is 's' dollars and the price of a child ticket is 'c' dollars.
From the information given, we can set up two equations:
- On the first day, the school sold 3 senior citizen tickets and 9 child tickets for a total of $75:
3s + 9c = 75
- On the second day, the school sold 8 senior citizen tickets and 5 child tickets for a total of $67:
8s + 5c = 67
We have a system of two equations with two unknowns (s and c). We can solve this system of equations to find the prices of the senior citizen ticket and the child ticket.
Using any method of solving simultaneous equations (substitution, elimination, or matrices), we can find that the price of a senior citizen ticket (s) is $9 and the price of a child ticket (c) is $4.
Therefore,
The price of one senior citizen ticket is $9 and the price of one child ticket is $4.
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