RAFAEL N. answered 07/11/19
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
Heather B.
asked 07/11/19RAFAEL N. answered 07/11/19
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 07/11/19
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 07/11/19
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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