How Do You Solve An Application That Results In A Dependent System Of Equations?

Some of the most interesting problems to solve are problems that lead to a system of linear equations where there are more variables than equations. Typically, these systems have many possible equations. Figuring out which of those solutions make sense and which do not is challenging.

Here is one such example

A restaurant owner orders a replacement set of knives, fork, and spoons. The box arrives containing 40 utensils and weighing 141.3 ounces (ignoring the weight of the box). A knife, fork, and spoon weigh 3.9 ounces, 3.6 ounces, and 3.0 ounces, respectively.

How many knives, forks and spoons are in the box?

Since we are being asked to find the number of knives, forks, and spoons, let’s make the following designations:

K: number of knives

F: number of forks

S: number of spoons

The first thing to notice is that you are given a total number of utensils (40) and a total weight for the utensils (141.3 ounces). These are the prime candidates for writing out the equations.

Starting with

Total number of utensils = 40

It is fairly obvious that

K + F + S = 40

Starting with

Total weight of utensils = 141.3

We can deduce that the individual weights are

Weight of knives = 3.9 K

Weight of forks = 3.6 F

Weight of spoons = 3.0 S

So

3.9 K + 3.6 F + 3.0 S = 141.3

Now what? Your experience in these sections probably tells you that you need another equation in the three unknowns to be able to solve for K, F, and S. This is true if there is a unique solution to this problem. But this problem actually has many possible solutions (ie. There are many ways to have 40 utensils that weigh 141.3 ounces).

So let’s simply solve the problem as is with Gauss-Jordan elimination:

K + F + S = 40

3.9 K + 3.6 F + 3.0 S = 141.3

Start by converting this to an augmented matrix:

To put into row echelon form, let’s multiply the first row by -3.9 and add it to the second row. Put the result in row 2:

To make the first nonzero number in the second row a 1, multiply the entire row by 1/-0.3 and replace the second row with that result:

This is now in row echelon form. It is almost as if there was a third row in the matrix, but it is all zeros.

With one more row operation, we can put the system in reduced row echelon form.

Converting this back to a system of equations, we get

K – 2S = -9

F + 3S = 49

To get our solutions, the first equation for K and the second equation for F:

K = 2S – 9

F = -3S+49

This is pretty easy to do since the system was in reduced row echelon form.

The variable S can be any value that makes sense for the problem. For instance, we certainly know that S (the number of spoons) should be non-negative integers like S = 0, 1, 2, … If we make up a table, we can make some interesting observations:

We need the numbers of each type of utensil to be positive…to do that we’ll require the number of spoons to be S = 5, 6, 7, …, 16. So even though the system has an infinite number of solutions, only certain ones make sense. In other words, any ordered triple of the form (K, F, S) = (2S – 9, -3S + 49, S) as long as S is an integer from 5 to 16.

How Do You Write Out the Solutions to a Dependent System of Linear Equations?

Suppose you are solving a system such as

2x + 3y = 3

4x + 6y = 6

Solving this system with substation or elimination leads to 0 = 0. This is a signal that there are an infinite number of solutions. This does not mean that ANY ordered pairs will solve the system. Only certain combinations of x and y will work. You need a way of finding any of those solutions. There are two ways to do this.

Method 1: Start with one of the equations (it does not matter which one) like 2x + 3y = 3. Solve this equation for x: 2x = -3y + 3 and then

If you have a value for y, this gives you a corresponding value for x. For instance, if y = 1, the corresponding x value is x = 0. This gives us one possible solution, (0, 1). If y = -1, then x = 3 giving us (3, -1). In general, we can write all solutions out as

Picking any value for y will give you a corresponding value for x which solves the system.

Method 2: What if we were to take 2x + 3y = 3 and solve for y. In this case we would get . If we were to pick values for x, we get corresponding values for x. For instance, if x = 0 we get y = 1 or the ordered pair (0, 1). Notice that this is one of the same ordered pairs as in Method 1. Let’s try another value for x, x = 3. When we put this into   we get y = -1. This gives the same ordered pair, (3, -1), as Method 1. In general, we can write out all possible solutions as

Both ways of writing the solution give the same ordered pairs. In Method 1, you pick a value for y and find the corresponding x value. In Method 2, you pick a value for x and find the corresponding y value. Since the values you pick can be anything, this gives the infinite number of ordered pairs that solve the system.

How Can You Model Data With A System of Equations (Continued)?

In an earlier FAQ, I mentioned that there was a second strategy for solving the Sony Math Problem. Recall the basic problem:

In December of 2014, Sony released the movie The Interview online after threats to theaters cancelled the debut in theaters. As originally reported in Wall Street Journal, the sales figures reported in January contained an interesting math problem appropriate for algebra students.

The following January, Sony reported sales of $31 million from the sales and rentals of The Interview. They sold the movies online for $15 and rented through various sites for $6. If there were 4.3 million transactions, how many of the transaction were sales of the movie and how many of the transactions were rentals?

Continue reading “How Can You Model Data With A System of Equations (Continued)?”

How Can You Model Data With A System of Equations?

In December of 2014, Sony released the movie The Interview online after threats to theaters cancelled the debut in theaters. As originally reported in Wall Street Journal, the sales figures reported in January contained an interesting math problem appropriate for algebra students.

The following January, Sony reported sales of $31 million from the sales and rentals of The Interview. They sold the movies online for $15 and rented through various sites for $6. If there were 4.3 million transactions, how many of the transaction were sales of the movie and how many of the transactions were rentals?

Continue reading “How Can You Model Data With A System of Equations?”

How Do You Model a Stock Portfolio Containing Two Stocks?

To model a simple stock portfolio with two stocks, we’ll write down a system of two equations in two variables. We hope to find a unique solution to this system, so let’s make sure we understand two key ideas.

  • We need two variables.
  • We need two equations.

Why are these important?

Two Variables?

The variables represent the two unknown quantities we are looking for. Since we want to know how much two invest in each stock in a tow stock portfolio, the two variables will represent the amounts of money invested in each stock. If we had more stocks in the portfolio, we would need more variables to correspond to.

Two Equations?

If we hope to solve our system of linear equations for a unique solution, the number of equations must match the number of variables. This assumes that one of these equations is not redundant. For this model, we’ll get our equations from two pieces of information, the total amount invested in the portfolio and the total return desired.

For a larger portfolio, we would need more equations to specify a unique solution. In that case we would need more information such as an average beta for the portfolio.

Write Out the System

For this portfolio, we will model two securities:

Based on data from the end of January 2016, we know the following information.

Security  Annual Dividend Yield  Beta
Tootsie 1.16% 0.7
Diebold 4.45% 1.51

 

Our goal for this example is to invest a total of $50,000 with a total dividend return of 3%. This is attainable since one security in the portfolio has a higher yield and the other a lower yield. It would be impossible to combine the stocks in a portfolio to get a total yield higher that the highest yielding stock of lower than the lowest yielding stock.

Start with your variables. I’ll call mine x1 and x2 and describe them as

x1: amount invested in Tootsie

x2: amount invested in Diebold

Once you understand what these are, it is easy to use the information in the problem to write out the two equations.

Total Amount Invested Is $50,000

We can start to get mathematical by writing

Total Amount Invested = 50,000

To finish the equation, we need to write the left side of the equation in terms of the variables. A “total” indicates addition so write

x1 + x2 = 50,000

Total Dividend Return is 3%

If the total dividend return needs to be 3% of $50,000, we need a total of

3% of $50,000 = (0.03)(50,000) = 1500

This total dividend will come from the dividend on the Tootsie stock,

1.16% of the amount invested x1 = 0.0116 x1

and the dividend on the Diebold stock,

4.45% of the amount invested x2 = 0.0445 x2

So if the total dividend from the portfolio is $1500 and this is the sum of the dividend from each stock in the portfolio,

0.0116 x1 + 0.0445 x2 = 1500

Model for a Two Stock Portfolio

Combining the two equations together gives a system of two linear equations in two variable,

x1 + x2 = 50,000

0.0116 x1 + 0.0445 x2 = 1500

We can solve these graphically or algebraically. If we use the substitution method and solve them graphically, solve for x1 in the first equation to give

x1 = 50,000 – x2

Putting this into the second equation leads to

0.0116 (50,000 – x2)+ 0.0445 x2 = 1500

        580 – .0116 x2 + 0.0445 x2 = 1500

                                 0.0329 x2 = 920

                                                       x2 ≈ 27,963.53

If $27,963.53 is invested in Diebold, then x1 ≈ 50,000 – 27,963.53 or $22,036.47 must be invested in Tootsie.

The sum of these amounts is $50,000 as desired and the total dividend is

0.0116(22036.47) + 0.0445(27963.53) ≈ 1500

as it should be.