The quantity of each type of seats sold are as follows:
- Movie and a dinner seat = 200
According to the question,there are four times as many 3D, x seats as Dinner and a Movie, y seats.
That is, x = 4y
Also, total seats
= (x) + (y) + (z) = 3000...….............eqn(1)
Also, If the theater brings in $53,000 when tickets to all 3000 seats are sold.
- 20x + 35y + 15z = 53000...........eqn(2)
By substituting 4y for x in equations 1 and 2; we have;
<em>5y + z = 3000</em>..…........eqn(3) and
<em>115y + 15z = 53000</em>.........eqn(4)
By solving equations 3 and 4 simultaneously; we have;
y = 200 and z = 2000
and since x = 4y
x = 800
The quantity of each type of seats sold is as follows:
- Movie and a dinner seat = 200
Read more:
brainly.com/question/12413726
I’m not sure if this is right I hope it helps but you weren’t five to the right with the negative one to the four so you have to go five to the left making it -6 for the first coordinate and then he went 13 to the left for the second coordinate so you have to go 13 to the right 6+13 is 19. So your answer is -6, 19 hope this helps!
