Answer :
x = premium seats
y = regular seats
You need two equations: (x + y - 10 = 1200) and (30x + 20y = 30,180)
1.) Isolate one variable
y = 1,210 - x
2.) Plug the new equation into the second ORIGINAL equation
30x +20(1,210 - x) = 30,180
3.) Ditribute
30x + 24,200 - 20x = 30,180
4.) Add like terms
10x + 24,200 = 30,180
5.) Subtract 24,000 from either side
10x = 6,180
6.) Solve for x
x = 618
7.) Plug in x to the equation of your choosing
618 + y - 10 = 1200
8.) Solve for y
608 + y = 1,200
y = 592
The theater sold 592 regular seats and 618 premium seats.
y = regular seats
You need two equations: (x + y - 10 = 1200) and (30x + 20y = 30,180)
1.) Isolate one variable
y = 1,210 - x
2.) Plug the new equation into the second ORIGINAL equation
30x +20(1,210 - x) = 30,180
3.) Ditribute
30x + 24,200 - 20x = 30,180
4.) Add like terms
10x + 24,200 = 30,180
5.) Subtract 24,000 from either side
10x = 6,180
6.) Solve for x
x = 618
7.) Plug in x to the equation of your choosing
618 + y - 10 = 1200
8.) Solve for y
608 + y = 1,200
y = 592
The theater sold 592 regular seats and 618 premium seats.
Answer:
Number of premium tickets sold = 638
Number of regular tickets sold = 552
Step-by-step explanation:
Let number of premium tickets sold be p and number of regular tickets be r.
Total number of seats = 1200
Ticket sales with 10 seats left on, that is
p + r = 1190 ----------------eqn 1
Cost of premium ticket = 30 $
Cost of regular ticket = 20 $
Money collected = 30180 $
Total money collected = 30 p + 20 r = 30180
3p + 2 r = 3018 ----------------eqn 2
eqn 1 x 2
2p + 2r = 2380 --------------------eqn 3
eqn 2 - eqn 3
p = 3018 - 2380 = 638
Substituting in eqn 1
638 + r = 1190
r = 552
Number of premium tickets sold = 638
Number of regular tickets sold = 552
