a($6) + s($4.5) + c($2) = $2,670

a + s – 270 = c

3s + 20 = a

We could find out the answer by using terms of a, s, or c, but let’s use s because we already know the value of a in terms of s (3s+20 = a)

c is equal to (a + s – 270)

To find out how much it’s equal to in terms of s we just replace a with it’s value in terms of s.

in other words:

c = (3s+20) + s – 270

or c= 4s – 250

Now that we know their values in terms of s we can plug them into the main problem and solve for s.

Once you have the value of s, you can easily find the other 2 values.

Adult price = 6

Senior price = 4.5

Child price = 2

(1) a + s = c + 270

(2) a = 3s + 20

(3) 6a + 4.5s + 2c = 2670

Three equations, three unknowns.

(1a) c = a + s – 270

Substitute into equation (3)

6a + 4.5s + 2(a + s – 270) = 2670

12a + 9s + 4a + 4s – 1080 = 5340

(4) 16a + 13s = 6420

(2) a = 3s + 20, substitute into (4)

16(3s + 20) + 13s = 6420

48s + 320 + 13s = 6420

61s = 6100

s = 100

(2) a = 3s + 20 = 320

(1a) c = a + s – 270 = 100 + 320 – 270 = 150

Therefore, there were 320 adult tickets, 100 senior tickets and 150 child tickets sold.

(Total price = 6×320 + 4.5×100 + 150×2 = $2670)

senior: x

adult: 3x+20

child: x + 3x+20-270 = 4x -250

multiple the amount with the prices & set equal to total price:

x(4.5) + (3x+20)(6)+(4x-250)(2) = 2670

4.5x + 18x+120+8x-500 = 2670

30.5x – 380= 2670

senior: x=100

adult: 320

child: 150

i didnt check it so you should probably do that before accepting this

.. wow . hangfire just changed his answer from “three unknowns” to somehow finding the answer lols

A + S = C + 270

A= 3S + 20

6A + 4.50S + 2.00C = 2, 670

but if u use this method of system of equations it gets really complicated.. the guy beneath probably did it a better way