Often the most difficult part of solving a system of equations problem is writing out the system from a word description. In this FAQ, we look at a complicated sounding problem and form a system of linear equations from the problem.

Let’s look at the problem below.

A knitting shop orders yarn from three suppliers in Toronto, Montreal, and Victoria. One month the shop ordered a total of 109 boxes of yarn from these suppliers. The delivery costs were 77 dollars. 45 dollars. and 63 dollars per box for the orders from Toronto, Montreal, and Victoria respectively, with total delivery costs of 6355 dollars. The shop ordered the same amount from Toronto and Victoria. How many boxes were ordered from each supplier?

When beginning a problem involving systems of equations, we need to define the variables in the problem. In this problem, we are looking for the number of boxes ordered from each location. With this in mind, we define

T: number of boxes ordered from Toronto
M: number of boxes ordered from Montreal
V: number of boxes ordered from Victoria

There are several pieces of information we can use to write out equations. The information “the shop ordered a total of 109 boxes of yarn from these suppliers” gives us

This information is distinguished by the fact that it is a total involving the variables.

Let’s look for another piece of information involving a total. The problem also gives the total delivery costs, 6355 dollars. We can use this to write out another equation if we can write out the delivery costs to each supplier. If a single box costs 77 dollars to deliver to Toronto, T boxes will cost 77T dollars to deliver to Toronto. Following this pattern, we write out

77T: cost to deliver T boxes to Toronto
45M: cost to deliver M boxes to Montreal
63V: cost to deliver V boxes to Victoria

This means to shipping costs to each city is related to the total costs by

The final equation may be written using the information, “The shop ordered the same amount from Toronto and Victoria”. This tells us that

If we put all of these equations together, we get the system of linear equations,

Move the variables to the left side of the equation to prepare the system for solving. This gives the system,

The most common mistakes in solving problems with systems of linear equations is in setting up the problem in the first place. In the example below, note how the variables are defined carefully so that the difference between amounts of money invested is distinguished from amounts of interest.

Example Katherine Chong invests 10,000 dollars received from her grandmother in three ways. With one part, she buys US savings bonds at an interest rate of 2.5% per year. She uses the second part, which amounts to twice the first, to buy mutual funds that offer a return of 6% per year. She puts the rest of the money into a money market account paying 4.5% annual interest. The first year her investments bring a return of 470 dollars. How much did she invest in each way?

Write Out The Equations

For any problem like this, we want to determine what the variables are and what they represent. Since the question asks, “How much did she invest in each way?”, let’s define

B: amount invested in bonds

F: amount invested in mutual funds

M: amount invested in money market account

How do these amounts relate to the other key totals given in the problem?

A total of 10,000 dollars invested.

The amount of invested in mutual funds is twice that invested in savings bonds.

The total return from the investments is 470 dollars.

We need to use the three facts to write out three equations in the three variables. Let’s look at each of the three facts.

A total of $10,000 invested

Since the variables represent the amounts invested, this statement simply says that the sum of the variables is 10,000 or

B + F + M = 10000

This type of fact is usually the easiest to write as an equation.

The amount of invested in mutual funds is twice that invested in savings bonds

Let’s identify the variables in this statement:

If we now translate this to a mathematical equation, we get

F = 2B

The total return from the investments is $740

To calculate the return on an investment, we need to multiply the percent return as a decimal times the amount invested at that percent. For instance, if B dollars are invested in bonds returning 2.5% per year then the amount returned from that investment in a year is .025 B. If we apply this thinking to each investment, we get that

.025B + .06F + .045M = 470

Now that we have three equations in three unknowns, let’s write out the system:

Rewrite the second equation by subtracting 2B from both sides:

Solve for Variables

We solve this system so that it looks like

where ?? are different numbers. To put in this form, we’ll utilize row operations. We’ll begin by making sure the coefficient on B in the first equation is a 1. Since it is, we do not need to use a row operation to make it so.

Now we’ll eliminate B from the second and third equations using the first equation. Use the operations

2E1 + E2 → E2

-.025E1 + E3 → E3

Let’s examine the first carefully:

Now the second operation

Writing out the operations makes them easier to check later on. Rewriting the system after these operations leaves us with

Notice that the first equation has a 1 in front of the first variable and the same variable has been eliminated from the other equations.

Now let’s make the coefficient on the second variable in the second equation equal to 1 using ⅓E2 → E2. The system is now

To eliminate F from the third equation, carry out -.035E2 + E3 → E3. Carrying out this operation gives us

Now let’s eliminate F from the first equation with -1E2 + E1 → E1 . Now the system is

Notice that these last few steps put a 1 in front of the second variable in the second equation and eliminate F in the other equations.

The last step in putting the system in echelon form is to make the coefficient on the third variable in the third equation a 1. To do this, -300E3 → E3. After multiplying the last equation, we have

Now let’s eliminate M from the first and second equation,

-⅔E3 + E2 → E2

-⅓E3 + E1 → E1

This gives

This means $2000 must be invested in bonds, $4000 in mutual funds, and $4000 in the money market account. As it should, the amounts add to $10,000 and there is twice as much in mutual funds as bonds. The total annual return is

The solution meets all of the requirements of the problem so it must be correct (unless we misinterpreted those requirements).

Note that we could solve the original system with matrices also and end up with the same solution.

What if the problem had said something like

The return on mutual funds is twice the return on savings bonds.

instead of

The amount of invested in mutual funds is twice that invested in savings bonds

Our problem above dealt with amounts invested…this new wording indicates a relationship between the returns. The rate on mutual funds is 6% so the return would be .06F. THe rate on bonds is 2.5% so the return would be .025B. Then this information is written in equation form as

.06F = 2 (.025B)

In this case, this would replace the second equation in the system.

I am a big advocate of using tree diagrams to solve problems involving conditional probability. However, some of these problems are easier to do if we reconstruct a frequency table from the probabilities in the problem.

Problem A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes the blood pressure (high, low, normal) and their heartbeats (regular or irregular). She finds that

When we carry out probability problems, the first task is to identify the events in the problem. Often we can identify a few key events and then use compliments, “and”, or “or” . This can simplify and help us to represent the information pictorially.