RAFAEL N. answered • 12d

Solid 20 years experience in Finance and teaching math to kids

A + B = 6.000

A*28 + B*40 = 187.200

SO

A= 6.000 - B

IN THE SECOND EQUATION YOU ADD THE A VALUE

(6000-B)*28 + B*40 = 187.200

B= 1600

A=6000-1600 = 4400

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RAFAEL N. answered • 12d

Solid 20 years experience in Finance and teaching math to kids

A + B = 6.000

A*28 + B*40 = 187.200

SO

A= 6.000 - B

IN THE SECOND EQUATION YOU ADD THE A VALUE

(6000-B)*28 + B*40 = 187.200

B= 1600

A=6000-1600 = 4400

Denise G. answered • 12d

Algebra, College Algebra, Prealgebra, Precalculus, GED, ASVAB Tutor

This is a problem with 2 variables, 2 unknowns.

Let x = Number of $28 tickets

Let y = Number of $40 tickets

Since we know there are 6000 seats in the theater

x+y=6000

Since we know we want the revenue to be 187200

28x+40y=187200

These are the 2 equations. You can solve by elimination, substitution or graphing. Either method would give you the solution:

x=4400

y=1600

**There were 4400 $28 tickets sold and 1500 $40 tickets sold.**

Mark H. answered • 12d

Tutoring in Math and Science at all levels

you have 2 unknowns and therefore you need 2 equations.

Assign X for the number of 28-dollar seats, and Y for the $40

From the first sentence:

X + Y = 6000

From the second sentence, the total revenue is $187,200. This is the sum of the revenue from both kinds of seats, so:

X*28 + Y*40 = 187,200

You now have 2 equations in X and Y. Solve by one of the standard methods.

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