Answer:
4200 tickets of [tex]\$28[/tex] and 1800 tickets of [tex]\$40[/tex] were sold
Step-by-step explanation:
Given:
Tickets are sold at price [tex]\$28[/tex] and [tex]\$40[/tex].
Let Number of tickets sold at price [tex]\$28[/tex] be [tex]x[/tex].
Let Number of tickets sold at price [tex]\$40[/tex] be [tex]y[/tex].
Theater has maximum capacity of 6000 seat.
Hence,
[tex]x+y=6000 \ \ \ \ equation \ 1[/tex]
Total revenue to be generated is [tex]\$189600[/tex]
[tex]\therefore 28x + 40y= 189600 \ \ \ \ equation \ 2[/tex]
Now Multiplying equation 1 by 40 we get,
[tex]x+y=6000\\40x+40y=240000\ \ \ \ equation \ 3[/tex]
Now Subtracting equation 2 by equation 3 we get,
[tex](40x+40y=240000)- (28x + 40y= 189600)\\12x=50400\\\\x=\frac{50400}{12}=4200[/tex]
Now substituting value of x in equation 1 we get,
[tex]x+y=6000\\4200+y=6000\\y=6000-4200= 1200[/tex]
Hence a total of 4200 tickets of [tex]\$28[/tex] and 1800 tickets of [tex]\$40[/tex] were sold
