The local theater has three types of seats for Broadway plays: main floor, balcony, and mezzanine. Main floor tickets are $48, balcony tickets are $45, and mezzanine tickets are $34. One particular night, sales totaled $65,140. There were 381 more main floor tickets sold than balcony and mezzanine tickets combined. The number of balcony tickets sold is 328 more than 2 times the number of mezzanine tickets sold. How many of each type of ticket were sold? I already have these equations: 48x+45y+34z=65140, x=y+z+381, y=2x+328, but I can't figure out how to substitute them.

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.

Step-by-step explanation:

step-by-step explanation: