Question 189537
Let a = adult tickets

Let c = children tickets

Then a + c = 400....This is one equation.

The second equation is 25a + 15c = 7200.

We have a system of linear equations in two variables.

a + c = 400
25a + 15c = 7200

Can you solve it now?

I suggest using the substitution method which far easier than any other method.

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I got your reply.
Let a = adult tickets 
Let c = children tickets 
Then a + c = 400....This is one equation. 
The second equation is 25a + 15c = 7200. 
We have a system of linear equations in two variables. 


a + c = 400...Equation A
25a + 15c = 7200 


I will solve for a in Equation A.

 

a = 400 - c

 

I will now plug 400 - c into Equation B for a.

 

25(400 - c) + 15c = 7200

 

10000 - 25c + 15c = 7200

 

-10c = 7200 - 10000

 

-10c = -2800

 

c = -2800/-10

 

c = 280 tickets

 

The number of children tickets is 280.

 

To find the number of adult tickets, we plug 280 for c into EITHER Equation A or Equation B (your choice).

 

I will use Equation A.

 

a + c = 400

 

a + 280 = 400

 

a = 400 - 280

 

a = 120

 

The number of adult tickets is 120.

 

Understand?

NOTE: The number of children tickets cannot be 250.  The correct answer is 280.

HERE IS THE PROVE:

adult tickets = 120 x $25 each = $3000

children tickets = 280 x $15 each = $4200

Together = $3000 + $4200 = $7200