1. Find the current population of the United States and round to the nearest million.

Find the current national debt total and round to the nearest trillion.

Based on these figures, calculate the approximate amount of debt per citizen.

Debt = 1000

Pop = 100

Debt/Pop=1000/100=10 per citizen

2. Assume that population growth is a linear function that grows at 0.9% per year, and

assume that debt growth is a linear function that grows at 13% per year.

Find the linear equation for population in slope-intercept form.

Find the linear equation for debt in slope-intercept form.

3. Graph the two linear equations in separate graphs. Label the values of the x and y axes.

Show x from 0 (current year) to 30 (30 years from now). Determine what the population

and national debt will be 30 years from now. Calculate what the debt per citizen will be in

30 years.

4. Graph the function that shows the amount of debt per citizen as a function of time. Is this

a linear function? What will the amount of debt per citizen be in 20 years?

5. If the current compound interest rate on our national debt is 8% (compounded annually),

what will the national debt be in 5, 10, 20 years? Show equation.

6. At what point does the compound interest expense double the amount of the first year?

Solve algebraically.

7. Given the linear function for population growth (in response #2), if we wanted the $ debt

per person ratio to increase at a constant rate of $2000 per year, what would the model

for the national debt need to look like for this to happen? Is it linear or quadratic? Explain

your reasoning.

College Algebra: Final Project

Current Population= 317 million from http://www.census.gov/popclock/ on 12/9/13

Current Debt= 17 trillion from http://www.usdebtclock.org/ on 12/9/13

$17,000,000,000,000/$317,000,000=$53,627.7603

Debt per citizen=$53,627.76

#1

Slope intercept equation for population: 0.9(#ofyears)+population=growth

9/10x+317=y

Slope intercept equation for debt: 0.13(#ofyears)+debt=growth

13/100x+17=y

#2

Population in 30yrs= 344 million

National Debt in 30yrs= 20.9 trillion

Debt per citizen in 30yrs= $60,755.81

**#3**

#5

5 yrs = 17(1*0.08/1)^(1)5= 25 trillion

10yrs = 17(1*0.08/1)^(1)10= 36.7 trillion

20 yrs = 17(1*0.08/1)^(1)20= 79.25 trillion

#6

#7

17(1*0.08/1)^(1)1

Which means in the first year the debt will accrue 1.36 trillion dollars of interest.

17(1*0.08/1)^y=19.72

So it will take approximately 1.93 years to double the expense of the first year.

This equation

y=2,000x+53,627.76

Using this model the national debt would increase by a total of $2,000 per citizen per year. This would be a linear equation because it is increasing at a constant rate over time which would be a straight line. If we had additional data the equation could potential change into a non- linear graph.

Part B

For this part, we will look at equations with two variables in a business setting.

Suppose a company’s profit was 15% lower in June than in May. The total profit for both

months is $32,375.

8. Declare what each of your variables will represent.

9. What are the two equations that will be used to solve this problem?

10. Find the profit for each month. Show all your work and provide justifications or explanations

for each step.

#8

**#9**

**#10**

J= June Profit

M= May Profit

**J+M=$32,375**

J=M-0.15M

J=M-0.15M

Plug the second equation in the first to find May's profit.

(M-0.15M)+M=$32,375

0.85M+M=$32,375

1.85M=$32,375

Divide both sides by 1.85

M=$17,500

Then plug M into the first equation.

J=$17,500-0.15($17,500)

J=$17,500-$2,625

J=$14,875

Lastly if you plug both M and J into either equation you will get $32,375.

Part C

Identify three situations in your life in which you could use algebra to solve real and important

questions or problems. Work situations are ideal, but any real situation in your life is

appropriate. Describe what types of equations or operations you would use to solve the

problems, and explain why algebra is helpful.

Is this a linear function? We assume that is it is a linear function since we have not collected or acquired additional data to produce a non-linear function.

Current debt over population: 17/317

f(20)=0.13x+17/g(20)=0.9x+317

Debt over population in 20 yrs=19.6/335

Debt per citizen in 20yrs= $58,507.46

Used (19.6, 335) (17, 317) to find the slope to graph.

#4

I have $58,000 saved for retirement. Assuming I will make 9% interest annually on my money, how l many years will I need to save $3 million dollars?

SITUATION #1

SITUATION #2

I would like to pay off my mortgage early. I owe $143,000 at 2.87% interest. The payment is $1237 per month with $750 of that going to the principal. How much additional do I need to pay to the principal each month to pay my mortgage off in 5 years?

SITUATION #3

I started my own business. I am making scarves. Each scarf costs $1.10 to make and I sell them for $15.30. In order to make a $30,000 dollar profit the first year, how many scarves would I need to sell in that year?

Compound Interest Formula:A=P(1+r/n)^nt

A=FinalAmount

P =Principle (starting balance)

r =Interest rate (as a decimal)

n=number of compounds per year

t=time (in years)

Used this formula for both #5 and #6

Once again I can apply the compound interest formula.

$3,000,000=$58,000(1+0.09/1)^1(y)

Solving this for y.

It will take approximately 45.8 years to accrue $3 million dollars without adding any additional funds.

(15.30-1.10)s=$30,000

14.20s=$30,000

s=2113

Therefore to achieve the profit goal I would need to sell 2113 scarves.

$143,000(1+0.0287/1)^(1)(5)=$164,732.67

Based on this calcuation over the next 5 years the loan will accrue an additional $21,72.67

Then we take the $750 per month and in five years that will pay $45,000 off the principal. Therefore we will need to pay an additional $119,732.67 off.

That means that over the 5 years we would need to pay an additional $1995.54 per month toward principal.

I used the compound interest formula again for situation #1. Algebra is helpful because it simplifies the work that needs to be done to calculate the problem by having a formula rather than needing to create your own.

In situation #3 I used a simple linear equation. Sometimes we think algebra always have extremely complex ways of solving problems but not always. Algebra can be a simple way to come up with a solution for a simple problem.

For situation #2 I used the compound interest formula along with some more basic linear equations with simple math like division and subtraction. Algebra makes what once was a confusing problem into a easily solved problem. Now I know how much I need every month to pay off my mortgage in 5 years.

or y=9/10x+317

or y=13/100x+17