Multiple variable world problem. PLEASE HELP I CANT FIGURE IT OUT!!!!!!!

Question

Seventy more student tickets than adult tickets were sold for the school Drama production. Student tickets cost $5, adult tickets cost $8, and a total of $3210 was collected. How many of each kind of ticket were sold?

Answer 1

8a +5s = 3210

a = s-70

8(s-70) +5s = 3210

8s -560 +5s = 3210

13s = 3210+560 = 3770

s = 3770/13 = 290 student tickets

a = s-70 = 220 adult tickets

get integer solutions that's a strong clue you did the math right

8(220) +5(290) = 1760 +1450 = 3210 they work, answer checks out

Answer 2

Let A = # adult tickets sold

S = # student tickets sold

70 more student tickets than adult tickets sold translates into algebra this way:

S = A + 70 (the number of student tickets sold IS 70 more then adult)

This can also be written as -A + S = 70.

As far as cost goes:

8A + 5S = 3210.

Now you have two equations with two unknowns. You can use substitution or addition/elimination methods to solve the system. Since we already have S = A + 70. I would just use substitution, replacing the S in the 2nd equation with A + 70:

So:

8A + 5(A+70) = 3210. (one equation, one unknown). Remove parenthesis (distributive property), combine similar terms, isolate the term containing A (subtraction property for equality), then divide both sides by the numerical coefficient of A (division property for equality). Once you have the value for A, substitute into the 1st equation to determine S.

Hopefully the answers are positive integers, since you cannot purchase a fraction of a ticket.