The local theater has three types of seats for broadway plays: main floor, balcony, and mezzanine. main floor tickets are $59,
balcony tickets are $50, and mezzanine tickets are $40. one particular night, sales totaled $73,785. there were 435 more main floor tickets sold than balcony and mezzanine tickets combined. the number of balcony tickets sold is 78 more than 33 times the number of mezzanine tickets sold. how many of each type of ticket were sold?
In this problem, you have three unknowns: broadway tickets for the (1) main floor, (2) balcony and (3) mezzanine. To solve algebraic equations, the total number of unknowns must be the same with the number of independent equations. With that, let's formulate these three equations.
Let the 3 unknowns be
x = number of tickets for the main floor y = number of tickets for the balcony z = number of tickets for the mezzanine
The independent equations are:
59x+50y+40z=73,785 --> eq 1 x = y+z+435 --> eq 2 y = 78 + 33 z ---> eq 3
Solve the equations simultaneously. Substitute eq3 to eq 2
x = 78+33z+z+435 x = 513+34z --> eq 4
Substitute eq 4 and eq 3 to eq 1
59(513+34z)+50( 78 + 33 z)+40z=73,785
Solve for z using the calculator: z = 11 Use eq 4 to find x: x = 877 Use eq 2 to find y: y = 432
There were 877 main floor tickets, 432 balcony tickets and 11 mezzanine tickets sold.
