# Math Help - help needed hope this belongs in this forum

1. ## help needed hope this belongs in this forum

not sure if enough information is given but here goes

Movie theatre

2013
number of people attending in 2013 = x1
avg ticket price = y1
2013 revenues x1*y1 = 30.6Million

2014
number of people attending in 2014 = x2
avg ticket price = y2
2014 revenues x2* y2 = 28Million

Revenue difference Difference 2014 -2013 =- 2.6 million
avg ticket price decreased 1.2% in 2014 which contributed 340,000 to the 2.6 M decline
attendance declined 7.1 in 2014 from 2013

Solve for x1 and x2

Thanks all for the help -- my brain is fried trying to solve this so I know 2 equations 2 unknowns but if you could help w/ the calculation I would be eternally grateful. I posted in advance forum also but will delete if someone can help me here

2. ## Re: help needed hope this belongs in this forum

Originally Posted by juggler
not sure if enough information is given but here goes

Movie theatre

2013
number of people attending in 2013 = x1
avg ticket price = y1
2013 revenues x1*y1 = 30.6Million

2014
number of people attending in 2014 = x2
avg ticket price = y2
2014 revenues x2* y2 = 28Million

Revenue difference Difference 2014 -2013 =- 2.6 million
avg ticket price decreased 1.2% in 2014 which contributed 340,000 to the 2.6 M decline
attendance declined 7.1 in 2014 from 2013

Solve for x1 and x2

Thanks all for the help -- my brain is fried trying to solve this so I know 2 equations 2 unknowns but if you could help w/ the calculation I would be eternally grateful. I posted in advance forum also but will delete if someone can help me here
attendance declined 7.1 ... what? from 2013, percent? million?

3. ## Re: help needed hope this belongs in this forum

attendance declined by 7.1% from 2014 to 2013 -- so Im guessing x1= (1-7.1%)* x2

4. ## Re: help needed hope this belongs in this forum

Originally Posted by juggler
attendance declined by 7.1% from 2014 to 2013 -- so Im guessing x1= (1-7.1%)* x2
I'm not getting any solutions. I also wonder why they include the 340,000.

5. ## Re: help needed hope this belongs in this forum

I started to help on this same question posted under advanced algebra.

The student did not say what he was studying so it is a little tricky trying to answer. The main point is that there are four unknowns and so a solution requires four equations, but sufficient information is available. The way the problem is worded it may be designed as an example of something that accountants call analysis of variance.

6. ## Re: help needed hope this belongs in this forum

There are in fact five equations, but I am having trouble showing that four are independent. I'll look at it more tomorrow.

7. ## Re: help needed hope this belongs in this forum

I find the subscript notation being used in this problem not as clear as possible.

REVISED NAMING

$x = 2013\ attendance.$

$x + u = 2014\ attendance.$

$y = 2013\ average\ price.$

$y + v = 2014\ average\ price.$

GIVENS

$xy = 30.60.$

$(x + u)(y + v) = 28.00 \implies xy + vx + uy + uv = 28 \implies vx + uy + uv = 28- xy = 28 - 30.6 = -\ 2.6.$

$u = -\ 0.071x.$

$v = -\ 0.012y.$

$vx = -\ 0.340.$

5 equations, 4 unknowns. Question: is system soluble?

$vx + uy + uv = -\ 2.6 \implies -\ 2.6 = (-\ 0.012y)x + (-\ 0.071x)y + (-\ 0.071x)(-\ 0.012y) = xy(0.000852 -0.071 -0.012) = -\ 0.082148xy.$

Using the first four equations we can reduce the system to two equations in two unknowns. Unfortunately, those two equations must be either dependent or inconsistent. If they are dependent, all is not lost because we have a fifth equation.

$-\ 2.6 = -\ 0.082148xy \implies xy = \dfrac{-\ 2.6}{-\ 0.082148} \approx 31.65 \ne 30.6.$ They are in fact inconsistent. The problem is insoluble.