Let us call x the number of $25 seats and y $40 seats, then we know that there are in total 5000 seats in a theatre; therefore, we have the equation
[tex]x+y=5000[/tex]Also, after selling this many tickets the total revenue should be $149, 000; therefore, we get the equation
[tex]25x+40y=149,0000[/tex]Hence, we have a system of equations with two equations and two unknowns.
We solve this system by substitution.
First, we solve for x in the first equation to get
[tex]x=5000-y[/tex]we then put this into the second equation to get
[tex]25(5000-y)+40y=149,000[/tex][tex]\rightarrow125,000-25y+40y=149,000[/tex][tex]\rightarrow125,000+15y=149,000[/tex][tex]\rightarrow15y=24,000[/tex][tex]\therefore y=1600.[/tex]Now that we have y, we now solve for x to get:
[tex]x=5000-1600[/tex][tex]x=3400.[/tex]Hence x = 3400 and y = 1600 and the correct statements are as follows.
The number of tickets for sale at $25 should be 3400.
The number of tickets for sale at $40 should be 1600.
