Algebra 1B!!!! Word question..

At the school play on Friday night, 120 tickets were sold. Student tickets cost $5 each and Adult tickets cost $7 each. The total sales for the evening were $670. How many of each type of ticket were sold?

x = 5 dollar tckets
y = 7 dollar tickets

x + y = 120

5x + 7y = 670

- 5(x + y = 120)

- 5x - 5y = - 600
5x + 7y = 670

2y = 70

y = 35

so

x + y = 120

x + 35 = 120

x = 85

Hey :)

Let's say the number of student tickets is "S".
and the number of adult tickets is "A".

Since 120 tickets were sold in total,
A + S = 120

That's our first equation.

Each Student ticket gives us $5 and each adult ticket gives us $7.
If we had sold "S" student tickets, we would have gotten 5S dollars.
Same way, if we sold "A" adult tickets, we would get 7A dollars.

So, the total amount of money we get is 5S + 7A.

And, we're given that this total amount is $670.

So, we've actually set up 2 equations for ourselves.

A + S = 120 ----- (1)
7A + 5S = 670 ----- (2)

Now, let's solve simultaneously.

We multiply the whole of equation (1) by 7.

(1) x 7: 7A + 7S = 840 ----- (3)

Since both equations (2) and (3) have 7A, we can subtract the 2 equations and 'get rid of' the A-term.

(3) - (2): 2S = 840 - 670
2S = 170
S = 170 / 2 = 85.

Since we now figured out that 85 student tickets were sold, we can easily say that 120 - 85 = 35 Adult tickets were sold :)

Hope that helps :)
- Urmila Sairam

Let students tickets sold be x and adult tickets sold be y.

So, x + y = 120 and 5x + 7y = 670.

Thus, x = 85 and y = 35.

let F + S = 120 # of 5 and & 7 doller tickets sold , then

5F+7S = $670

Solve by substition