System of 3 Equations Word Problem Examples

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Solving a System of Equations with No Solution

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
Your next lesson will play in 10 seconds
  • 0:04 System of Three Equations
  • 0:34 Tropical Punch Word Problem
  • 4:21 College Football Word Problem
  • 7:00 Lesson Summary
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Speed Speed

Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Michael Quist

Michael has taught college-level mathematics and sociology; high school math, history, science, and speech/drama; and has a doctorate in education.

Systems of three equations allow us to resolve situations where there are three specific unknowns. We'll explore two different word problem examples of such systems and look at how to set up and solve those systems of equations.

System of Three Equations

A system of three equations is a set of three equations that all relate to a given situation and all share the same variables, or unknowns, in that situation. A system of three equations can be used to solve a problem where there are three unknowns and enough information to make three equations. There are several different methods for solving systems of three equations, and in this lesson we are going to use the two most popular: substitution and elimination. The following examples illustrate word problems involving three equations and three unknowns.

Tropical Punch Word Problem

Suppose you want to make a certain kind of tropical punch, using bananas, oranges, and papayas. You don't know how many of each to put in the punch, but you know that there are seven pieces of fruit in the mix, and there are twice as many oranges as bananas. You also know that the seven pieces of fruit cost $5.25, where bananas cost $.50 each, oranges cost $.75 each, and papayas cost $1.25 each.

Since you do not know how many of each type of fruit to put in the drink, those are the unknowns for the system of three equations. The most common variables used are x, y, and z. However, to avoid confusion, we are going to let b represent the number of bananas used in the drink, o represent the number of oranges, and p represent the number of papayas. Let's use these variables and turn the information in the word problem into mathematical equations:

Equation 1: b + o + p = 7 (total number of pieces of fruit)

Equation 2: 0.5b + 0.75o + 1.25p = 5.25 (total cost of the fruit, where the number of each piece, indicated by the variable, is multiplied by its individual cost)

Equation 3: o = 2b (twice as many oranges as bananas in the mix)

Since we know that o = 2b, we can use substitution. Substitution is when you have a known value for one variable and replace that known value in the other equations. In this case, we can substitute 2b for o in the other two equations. This reduces a three-equation problem to a two-equation problem - much simpler to solve!

b + 2b + p = 7 simplifies to become

3b + p = 7

.5b + .75(2b) + 1.25p = 5.25 simplifies to become

2b + 1.25p = 5.25

We've now managed to remove the o variable from the equations, which helps us move toward a solution, and we can now use the process called elimination, to combine the remaining two equations and temporarily remove the p variable.

When using the elimination method, the goal is to find a way to remove, or eliminate, a variable by adding the two equations. If you are lucky, a variable in each equation will be the opposite of each other and automatically cancel out when adding the equations together (for example, 3x and -3x would cancel out).

Unfortunately for you, in this example, life is not that easy. So, we have to create a way to get a variable to cancel out in each equation. If we multiply the 3b + p = 7 equation by -1.25, we can convert the p term into an opposite of the one in the 2b + 1.25p = 5.25 equation. Then, we can add the two equations together to eliminate the p term from the resulting equation.

(-1.25) (3b + p = 7) results in

-3.75b -1.25p = -8.75


-3.75b -1.25p = -8.75 to

2b + 1.25p = 5.25

We'll see that the p term cancels out, leaving a simple single-variable linear expression, which we may solve directly:

-1.75b = -3.5

b = 2

Once we have a value for b, the rest is easy. We now go back to our original three equations and input our numbers in place of the variables. Since o = 2b, o must be 4, and since b + o + p = 7, then p must be equal to 1. The punch recipe requires two bananas, four oranges, and one papaya!

In the next example, we will use a three-equation system to determine college football statistics.

To unlock this lesson you must be a Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use

Become a member and start learning now.
Become a Member  Back
What teachers are saying about
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account