Equation question.

Question

Your school sold 456 tickets in a school play. An adult ticket(a) costs 3.50, while a student(n) costs 1$. Total ticket sales made it to 1131. Find each type of ticket sold.

Would the correct way to solve this be:

3.50a+ 1n=1131

a+n=456

then multiple to either cancel the a or n out

-3.50a-3.50n=-1596

-2.50n=-1596

n=638.4

a= 140.74....

Or

3.50a+1n=1131

a+n=456

times negative one

subtract

2.50a=675

a=270?

I don't understand what I am doing wrong

 

Answer 1

Let (a) equal the number of adult ticket sold.

Let (n) equal the number of student sold.

• Then the equation for the total number of tickets sold is
• a + n = 456 - Equation 1
• The total cost of tickets can be represented by the following equation
• $3.50a + $1n = $1131 - Equation 2
• Multiply Equation 1 by negative 1.
• -a - n = -456 add this to Equation 2.
• $3.5a + $1n = $1131

The result is 2.5a = 675

There a = 270. and n = 456 - 270 = 186